Frank Solutions and MCQ for Chapter 4 Current Electricity Class 10 Physics ICSE
Exercises
1. Tick (√) the correct choice in the following :
(a) In the circuit shown in fig. 1, the e.m.f. of the cell is 4 volts, the current flowing through the resistor is
(i) 2A/3
(ii) 2A
(iii) 3A/2
(iv) 4A/3
(v) 4 A
Answer
(i) 2/3 A
(b) When the length of a conductor is doubled the resistance of the conductor will
(i) Remain same
(ii) Double in value
(iii) Become four times
(iv) Become one fourth
(v) Become half
(i) Remain same
(ii) Double in value
(iii) Become four times
(iv) Become one fourth
(v) Become half
Answer
(ii) double in value
(ii) double in value
(c) Ohm's law is applicable to
(i) Discharge of electricity through gases
(ii) Diode value
(iii) All metallic conductors
(iv) Carbon compounds
(i) Discharge of electricity through gases
(ii) Diode value
(iii) All metallic conductors
(iv) Carbon compounds
Answer
(iv) ohmic conductors
(iv) ohmic conductors
(d) A 24 watt, 12 volt head lamp is fully lit by connecting it to a 12 volt battery. The working resistance of the lamp is
(i) 5 Ω
(ii) 4 Ω(i) 5 Ω
(iii) 1 Ω
(iv) 6 Ω
(v) 2 Ω
Answer
(iv) 6 Ω
(iv) 6 Ω
(e) Which of the following quantities has the same value for all electrical appliances connected to the ring main circuit in a house ?
(i) Power
(ii) Voltage
(iii) Resistance
(iv) Current
(v) Energy
(i) Power
(ii) Voltage
(iii) Resistance
(iv) Current
(v) Energy
Answer
(ii) voltage
(ii) voltage
(f) The outer casing of an electric iron is generally connected to earth in order to
(i) Protect the user from electric shock by short circuiting and consequently breaking the circuit.
(ii) Prevent the fuse from burning out.
(iii) Protect the iron.
(iv) Complete the electrical circuit to the iron.
(v) Allow the current to get away.
(i) Protect the user from electric shock by short circuiting and consequently breaking the circuit.
(ii) Prevent the fuse from burning out.
(iii) Protect the iron.
(iv) Complete the electrical circuit to the iron.
(v) Allow the current to get away.
Answer
(i) protect the user from electric shock by short circuiting and consequently breaking the circuit.
(i) protect the user from electric shock by short circuiting and consequently breaking the circuit.
(g) A three pin plug is connected to the lead for a 1 kW electric iron to be used on a 250 V supply. Which of the following statements is not correct ?
(i) The fuse should be fitted in the live wire.
(ii) The live wire is coloured brown.
(iii) A 13 ampere fuse is the most suitable rating to use.
(iv) The yellow and green wire should be connected to the earth pin.
(v) The blue wire should be connected to the neutral side of the mains.
(i) The fuse should be fitted in the live wire.
(ii) The live wire is coloured brown.
(iii) A 13 ampere fuse is the most suitable rating to use.
(iv) The yellow and green wire should be connected to the earth pin.
(v) The blue wire should be connected to the neutral side of the mains.
Answer
(iii) A 13 ampere fuse is the most suitable rating to use
2. Define the following:
(a) Potential difference
(b) (i) Coulomb
(ii) Ohm
(c) Electromotive force
(d) Semiconductors
(e) Super conductors.
(b) (i) Coulomb
(ii) Ohm
(c) Electromotive force
(d) Semiconductors
(e) Super conductors.
Answer
(a) Potential difference: The potential difference between two points may be defined as the work done in moving a unit positive charge from one point to the other.
(b) (i) Coulomb: It is the unit of charge.
(ii) Ohm: It is the unit of resistance. The resistance of a conductor is said to be 1 ohm, if 1 ampere current flows through it, when the potential difference across its ends is 1 volt.
(ii) Ohm: It is the unit of resistance. The resistance of a conductor is said to be 1 ohm, if 1 ampere current flows through it, when the potential difference across its ends is 1 volt.
(c) Electromotive force: When no current is drawn from a cell, when the cell is in open circuit, the potential difference between the terminals of the cell is called its electromotive force (or e.m.f.).
(d) Semiconductors: Substances whose resistance decreases with the increase in temperature are named as semiconductors.
Examples: manganin, constantan etc.
Examples: manganin, constantan etc.
(e) Superconductors: Substance whose resistance decreases tremendously with the decrease in temperature and reaches nearly zero around absolute zero temperature are named as superconductors.
Exampels: lead, tin etc.
Exampels: lead, tin etc.
3. State Ohm's law. What are its limitations?
Answer
According to Ohm's law, the current flowing in a conductor is directly proportional to the potential difference across its ends provided the physical conditions and temperature of conductor remains constant.
Limitations of Ohm's law:
Limitations of Ohm's law:
- Ohm's law does not apply to conductors such as diode, radio valves, metal rectifiers, where electricity passes through gases.
- Ohm's law is applicable only when the physical conditions remain constant.
- Ohm's law is applicable only when the temperature of the conductor is constant.
4. What are the factors on which the resistance of a conductor depends? Distinguish between conductors and insulators.
Answer
Factors on which the resistance of a conductor depends are:
- Nature of conductor: different materials have different concentration of free electrons and therefore resistance of a conductor depends on its material.
- Length of conductor: Resistance of a conductor is directly proportional to the length of a conductor.
- Area of cross-section of a conductor: Resistance of a conductor is inversely proportional to the area of cross-section of the uniform wire.
- Temperature of conductor: In general for metallic conductors, higher the temperature larger is the resistance.
5. Fig. 2 represents the circuit used for the verification of Ohm's law. Label the parts from A to F. state the function of each.
Answer
Functions:
(A) Cell- It provides the potential difference in the circuit.
(B) Key- It serves as a switch in the circuit. It supplies or cuts off current as required.
(C) Ammeter- It measures the current in the circuit.
(D) Rheostat- It helps to change the resistance of the circuit without changing its voltage.
(E) Resistor- It provides a constant resistance in the circuit.
(F) Voltmeter- It measure the potential drop across the resistor.
6. What is the combined resistance of each of the networks between A and B shown in fig. 3?
Answer
(a) Between A and B resistances 5Ω and 3Ω are connected in series.
This series combination of resistances 5Ω and 3Ω is connected in parallel with the resistance 8Ω
∴ total resistance between A and B is given as :
Parallel resistance of 9 Ω and 18 Ω is.
∴ total resistance between A and B is .
R = 6+2 = 8Ω
(c) The situation consists of three two ohm resistors connected in series between CEFD and their combination in parallel with the fourth 2 ohm resistor between C and D.
Therefore, series combination give, 2 + 2+ 2= 6 Ω.
This 6 Ω resistor is connected in parallel to the fourth 2 Ω resistor, therefore equivalent resistance between C and D,
Now, between A and B the resistance R and two 2 Ω resistor are connected in series.
Therefore, equivalent resistance between A and B is
7. A charge of 80 C flows in a conductor for 2 minutes.
(a) Calculate the current flowing through the conductor.
(b) If the current through a heater is 4 A what charge must be passing in 8 seconds?
(b) If the current through a heater is 4 A what charge must be passing in 8 seconds?
Answer
(a) Given, charge q = 80 C, time t = 2 minute = 120 s
Current, I = q/t = 80/120 = 0.67 A
(b) Given, current I = 4 A, time 8 s
Let q be the charge passing in 8s
Then, q =It = 4× 8 = 32 C
Current, I = q/t = 80/120 = 0.67 A
(b) Given, current I = 4 A, time 8 s
Let q be the charge passing in 8s
Then, q =It = 4× 8 = 32 C
8.The voltage across a 3Ω resistance is 6 V. How large is the current ? What is the resistance of a filament lamp when a voltage of 3 V across it causes a current of 0.5 A ?
Answer
If voltage, V = 6 volt and resistance R = 3Ω
The a/c to ohm's law, current I = V/R = 6/3 = 2 A
For filament lamp :
Given that, voltage = 3 V, current = 0.5 A.
let R be the resistance of the filament lamp;
then, R = V/I = 3/0.5 = 6 Ω
For filament lamp :
Given that, voltage = 3 V, current = 0.5 A.
let R be the resistance of the filament lamp;
then, R = V/I = 3/0.5 = 6 Ω
9. Two resistors P and Q of the same material and length but of different thickness are connected in parallel to a battery. The cross- sectional area of P is twice that of Q. What is the ratio of
(a) The resistance of P to the resistance of Q?
(b) The current in P to the current in Q?
(b) The current in P to the current in Q?
Answer
10. Show how would you connect three resistors, each of resistance 6Ω so that the combination has s resistance of (a) 9 Ω (b) 4Ω.
Answer
(a) To get an equivalent resistance of 9 Ω using three 6 Ω resistors, they should be connected in parallel as shown :
(b) To get an equivalent resistance of 4 Ω using three 6 Ω resistors, they should be connected in parallel as shown:
11. A battery of e.m.f. 3 V is connected in series with an ammeter , a 10Ω coil of wire and with a parallel combination of resistance of 3 Ω and 6 Ω. Draw a circuit diagram for the above arrangement. What is the
(a) Resistance of the parallel combination ?
(b) Reading on the ammeter ?
(c) Potential difference across the 3 Ω resistor ?
(d) Current flowing through each resistor ?
(a) Resistance of the parallel combination ?
(b) Reading on the ammeter ?
(c) Potential difference across the 3 Ω resistor ?
(d) Current flowing through each resistor ?
Answer
Circuit diagram :
12. Find the effective resistance in each of the following circuit diagrams (Fig. 4):
Answer
In the given network, between B and C the series combination of two 6Ω resistors is connected in parallel with the third 6Ω resistor.
∴ Resistance in parallel is :
Effective resistance between P and Q is
R = 10 + 30 + 20 = 60 Ω
(c) In the given network, the parallel combination of two 2Ω resistors is connected in series with the parallel combination of two 3Ω resistors.
∴ Resistance in parallel of two 2Ω resistors is :
The situation consists of three 10 ohm resistors connected in series between combination in parallel with the fourth 10 ohm resistor between C and D.
Therefore, series combination gives, 10+10+10 = 30 Ω.
This 30Ω resistor is connected in parallel to the fourth 10 Ω resistor, therefore, resistance between C and D,
Therefore, series combination gives, 10+ 10 +10 = 30 Ω
This 30 Ω resistor is connected in parallel to the fourth 10 Ω resistor, therefore resistance between C and D,
Now, between A and B the resistance R and two 10 Ω resistors are connect
Therefore, equivalent resistance between A and B is
In the path ACB, the series combination of two 2Ω resistor is connected in parallel with the third 2 Ω resistor.
∴ resistance in series = 2+ 2 = 4 Ω
This 4 Ω resistor is now connected in parallel with the 2Ω resistor.
13. Two cells each having an e.m.f. of 2V and an internal resistance of 2Ω are connected (a) In series, and (b) In parallel as shown in fig. 5. What is the current flowing through the circuit in each case ?
(a) Given, emf(e) of the battery = 2×2 = 4V,
Internal resistance r = 2×2 = 4 Ω
Resistance in the circuit, R = 4 Ω
Total resistance of the given series circuit = 4+4 = 8Ω
Internal resistance r = 2×2 = 4 Ω
Resistance in the circuit, R = 4 Ω
Total resistance of the given series circuit = 4+4 = 8Ω
14. In the circuit shown in fig. 6, find the reading of the ammeter A when the switch S is
(a) Opened.
(b) Closed.
(b) Closed.
Answer
(a) When the switch S is opened, resistance 60 Ω is not connected in the circuit and current flows through the 20 Ω resistor.
∴ Resistance of the circuit, when the switch is open :
R = 20 + 5 = 25 Ω
Given, voltage = 10 V
∴ Current through the circuit, I = V/R = 10/25 = 0.4 A
The reading of the ammeter is thus 0.4 A, when the switch is open.
(b) When the switch is closed, resistance 60Ω gets connected to the circuit.
Thus, in the circuit the parallel combination of 60 Ω and 20 Ω resistors is connected in series with the 5 Ω resistor.
∴ Resistance of the circuit, when the switch is open :
R = 20 + 5 = 25 Ω
Given, voltage = 10 V
∴ Current through the circuit, I = V/R = 10/25 = 0.4 A
The reading of the ammeter is thus 0.4 A, when the switch is open.
(b) When the switch is closed, resistance 60Ω gets connected to the circuit.
Thus, in the circuit the parallel combination of 60 Ω and 20 Ω resistors is connected in series with the 5 Ω resistor.
15. A current of 0.3 A flowing through a branch of 6 Ω resistors in a junction as shown in fig. 7. Calculate the
(a) P.d. across the junction XY,
(b) Current flowing through 9 Ω and 24 Ω.
(c) P.d. across 24 Ω resistor, and
(d) e.m.f. of the cell.
(a) P.d. across the junction XY,
(b) Current flowing through 9 Ω and 24 Ω.
(c) P.d. across 24 Ω resistor, and
(d) e.m.f. of the cell.
Answer
(a) Potential difference across XY = pot. diff. across 6 Ω resistor.
= 0.3 × 6= 1.8 V
(b) Potential difference across 9 Ω resistor = Pot. diff. across 6 Ω resistor = 1.8 V
∴ Current through 9 Ω resistor = 1.8/9 = 0.2 A
Current through 2.4 Ω resistor = total current in the circuit = 0.3 + 0.2 = 0.5 A
(c) p.d. across 2.4 Ω resistor = 0.5 ×2.4 = 1.2 V
(d) emf of cell = p.d. across 2.4 Ω resistor + pot. diff. across XY = 1.2 + 1.8 = 3.0 V
(b) Potential difference across 9 Ω resistor = Pot. diff. across 6 Ω resistor = 1.8 V
∴ Current through 9 Ω resistor = 1.8/9 = 0.2 A
Current through 2.4 Ω resistor = total current in the circuit = 0.3 + 0.2 = 0.5 A
(c) p.d. across 2.4 Ω resistor = 0.5 ×2.4 = 1.2 V
(d) emf of cell = p.d. across 2.4 Ω resistor + pot. diff. across XY = 1.2 + 1.8 = 3.0 V
16. The circuit diagram (Fig. 8) shows a battery of e.m.f. 6 volts and internal resistanc of 0.8 Ω connceted in series. Find the
(a) Current recorded by the ammeter,
(b) P.d. across the terminals of the resistor B,
(c) Current passing through each of the resistors, B, C and D, and
(d) P.d. across the terminals of the battery.
(a) Current recorded by the ammeter,
(b) P.d. across the terminals of the resistor B,
(c) Current passing through each of the resistors, B, C and D, and
(d) P.d. across the terminals of the battery.
Answer
Given, emf e = 6 V, internal resistance r = 0.8 Ω
In the given circuit, parallel combination of resistances 2 Ω and 3Ω is connected series with a 3 Ω resistance.
In the given circuit, parallel combination of resistances 2 Ω and 3Ω is connected series with a 3 Ω resistance.
17. Complete the following table. The first answer has been given as an example .
|
Quantity |
Unit |
|
(a) Electrical potential |
Volt |
Answer
|
Quantity |
Unit |
|
1. Electrical potential |
Volt |
18. The following appliances are to be used on 240 V supply. Calculate the current used by each and say which fuse, 2 amp, 5 amp or 13 amp should be incorporated with each, (i) A television rated at 150 W, (ii) An electric iron rated at 750 W, (iii) An immersion heater rated at 3000 W, (iv) A hair dryer rated at 500 W. How much will it cost to run the television set for 100 days for an average of 4 hours a day at 60 paise per unit?
Answer
We know that P = VI
∴ I = P/V
Given voltage supply = 240 volt
∴ I = P/V
Given voltage supply = 240 volt
19. (a) What is the purpose of fuse in an electrical circuit?
(b) What is the criteria for selecting a fuse wire?
(c) If a fuse contains a 5 A fuse wire and the voltage is 240 V, what is the maximum power (wattage) which may be taken from the circuit?
(d) What would be the danger involved in replacing a blown fuse with the one which would carry a large current?
(c) If a fuse contains a 5 A fuse wire and the voltage is 240 V, what is the maximum power (wattage) which may be taken from the circuit?
(d) What would be the danger involved in replacing a blown fuse with the one which would carry a large current?
Answer
(a) An electric fuse is a device which is used to limit the current in an electric circuit. The use of a fuse thus safeguards the circuit and appliances connected in that circuit from being damaged. It is a short piece of wire made of an alloy of lead and tin. If the current passing through the fuse exceeds the safeguard limit the heat produced melts the fuse and this breaks the circuit.
(b) A fuse wire should have high resistance and low melting point.
(c) Maximum power that can be taken, P = VI = 240 x S = 1200 W.
(c) Maximum power that can be taken, P = VI = 240 x S = 1200 W.
(d) Replacing a fuse with another fuse of higher rating would allow a large current to flow through the circuit or appliance to which the fuse is connected and thus, there is a chance of short-circuiting and the appliance may get damaged.
20. An electric kettle has a heating element which is rated at 2 kW when connected to a 250 V electric supply. Calculate the
(a) Current which flows in the element when it is connected to a 250 V supply,
(b) Resistance of the filament,
(c) Heat produced in 1 minute by the element,
(d) Cost of running the kettle for 10 minutes a day for 30 days of the month, at the rate of Rs. 3.00 per unit.
(b) Resistance of the filament,
(c) Heat produced in 1 minute by the element,
(d) Cost of running the kettle for 10 minutes a day for 30 days of the month, at the rate of Rs. 3.00 per unit.
Answer
21. Three fuses available are rated at 2 A, 10 A and 13 A. Which fuse would you chose for an electric wire rated at 3 kW, 250 V? Show how you arrived at your answer.
Answer
Given power P = 3 k, W = 3000 W, Voltage = 250 V
We know that, P = VI
∴ I = P/V = 3000/250 = 12 A
Thus a maximum of 12 A current can pass through the electric wire and a fuse of rating 13 A should be used with it.
We know that, P = VI
∴ I = P/V = 3000/250 = 12 A
Thus a maximum of 12 A current can pass through the electric wire and a fuse of rating 13 A should be used with it.
22. 60 joules of heat was dissipated in a resistor when 20 C flowed for 5 s. Calculate:
(a) P.d. across the resistor,
(b) Resistance of the resistor, and
(c) Average power dissipated in the resistor.
(b) Resistance of the resistor, and
(c) Average power dissipated in the resistor.
Answer
23. Fig. 9 shows a 3 pin plug with the cover removed. The electric cable connected to the plug contains 3 wires with colour coded insulation namely brown, blue and green.
(a) Identify eaFla. ch of the colour coded wires by stating to which of the terminals A, B or C in the diagram they should be connected.
(b) Identify the terminal through which no current passes in normal circumstances.
(c) What is the purpose of the earthed wire connected to an electric appliance like an electric heater? Describe how it works.
(b) Identify the terminal through which no current passes in normal circumstances.
(c) What is the purpose of the earthed wire connected to an electric appliance like an electric heater? Describe how it works.
Answer
(a) Brown wire or live wire should be connected to terminal C.
Blue wire or neutral wire should be connected to terminal B.
Green wire or earth wire should be connected to terminal A.
(b) No, current passes through the earth terminal i.e. terminal A in normal circumstances.
(c) The metal case of an electrical appliance is earthed so that in any case of accidental contact of live wire with the metallic body of the appliance, the earth wire would provide a safe and easy path for the electric charges to flow down to the earth which acts as very large sink. Thus, user is thereby protected from any fatal electric shock.
Blue wire or neutral wire should be connected to terminal B.
Green wire or earth wire should be connected to terminal A.
(b) No, current passes through the earth terminal i.e. terminal A in normal circumstances.
(c) The metal case of an electrical appliance is earthed so that in any case of accidental contact of live wire with the metallic body of the appliance, the earth wire would provide a safe and easy path for the electric charges to flow down to the earth which acts as very large sink. Thus, user is thereby protected from any fatal electric shock.
24. Explain the meaning of the term kilowatt hour.
Answer
'kilowatt-hour' is the commercial unit of electricity. One kilo-watt hour is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for 1 hour.
25. Describe the construction and working of a filament lamp. List the material used and state why they are suitable for their purpose in this lamp.
Answer
Construction and working of filament bulb
List of materials used:
Light bulbs have two metal contacts, which connect to the ends of an electrical circuit. The metal contacts are attached to two stiff wires, which are attached to a thin metal filament. The filament sits in the middle of the bulb, held up by a glass mount. The wires and the filament are housed in a glass bulb, which is filled with an inert gas, such as argon.
In a modern light bulb, inert gases, typically argon, greatly reduce this loss of tungsten. At extreme temperatures, the occasional tungsten atom vibrates enough to detach from the atoms around it and flies into the air resulting in its evaporation. In the presence of argon gas around it, the chances are that it will collide with an argon atom and bounce right back toward the filament, where it will rejoin the solid structure. Also since inert gases normally don't react with other elements, there is no chance of the elements combining in a combustion reaction.
List of materials used:
Light bulbs have two metal contacts, which connect to the ends of an electrical circuit. The metal contacts are attached to two stiff wires, which are attached to a thin metal filament. The filament sits in the middle of the bulb, held up by a glass mount. The wires and the filament are housed in a glass bulb, which is filled with an inert gas, such as argon.
- When the bulb is connected to a power supply, an electric current flows from one contact to the other, through the wires and the filament.
- As the electrons zip along through the filament, they are constantly bumping into the atoms that make up the filament. The energy of each impact vibrates an atom -- in other words, the current heats the atoms up.
- Metal atoms release mostly infrared light photons, which are invisible to the human eye. But if they are heated to a high enough level around 4,000 degrees Fahrenheit in the case of a light bulb they will emit a good deal of visible light.
In a modern light bulb, inert gases, typically argon, greatly reduce this loss of tungsten. At extreme temperatures, the occasional tungsten atom vibrates enough to detach from the atoms around it and flies into the air resulting in its evaporation. In the presence of argon gas around it, the chances are that it will collide with an argon atom and bounce right back toward the filament, where it will rejoin the solid structure. Also since inert gases normally don't react with other elements, there is no chance of the elements combining in a combustion reaction.
26. In a house there are 6 bulbs of 100 W each used for 4 hours a day, 4 bulbs of 60 W each used for 8 hours a day, an immersion heater of 2.5 kW used for 1 hour per day, and an electric iron of 800 W used for 2 hours per day. Calculate the cost of monthly electric bill for the month of April at the rate of 80 p per unit.
Answer
27. Name the material and type of wires for the following:
(a) Connecting wires for lighting
(b) Fuse wire
(c) Heating elements
(d) Connecting wires of a power line
(e) Earthing elements
(b) Fuse wire
(c) Heating elements
(d) Connecting wires of a power line
(e) Earthing elements
Answer
(a) Connecting wires: Materials having low resistance low resistivity and high melting point e.g. copper, aluminium.
(b) Fuse wire: Materials having high resistance and low melting point e.g. solder an alloy of lead and tin.
(c) Heating element: Materials having high resistivity and high melting point e.g. tungsten.
(d) Connecting wire of a power line: Materials having low resistance and non-corosive properties e.g. high tension wires.
(e) Earthing elements: Materials which are good conductors of electricity. Earthing elements are copper wire, copper plate, salt.
(b) Fuse wire: Materials having high resistance and low melting point e.g. solder an alloy of lead and tin.
(c) Heating element: Materials having high resistivity and high melting point e.g. tungsten.
(d) Connecting wire of a power line: Materials having low resistance and non-corosive properties e.g. high tension wires.
(e) Earthing elements: Materials which are good conductors of electricity. Earthing elements are copper wire, copper plate, salt.
28. Two bulbs are marked 60 W, 220 V and 60 W, 110 V respectively. Calculate the ratio of their resistance.
Answer
29. An electric oven is rated as 1.5 kW, 250 V. If it is connected to a 250 V mains, calculate the
(a) Current drawn,
(b) Energy consumed in 20 hours,
(c) Cost of energy consumed in 20 hours at the rate of Re. 1 per unit.
(b) Energy consumed in 20 hours,
(c) Cost of energy consumed in 20 hours at the rate of Re. 1 per unit.
Answer
Given, Power P = 1.5 kw, voltage V = 250 V
30. An auto lamp is joined to a battery of e.m.f. 4V and internal resistance 2.5 Ω steady current of 0.5 A flows through the circuit. Calculate the
(a) Total energy provided by battery in 10 minutes,
(b) Heat dissipated in the bulb in 10 minutes.
(a) Total energy provided by battery in 10 minutes,
(b) Heat dissipated in the bulb in 10 minutes.
Answer
Given, emf, e = 4 V, internal resistance r = 2.5 Ω, current I = 0.5A
(a) Energy provided by battery in 10 mins = Power x time = (VIt)= 4×0.5×20= 40 watt-hour
(b) Heat dissipated in the bulb in 10 minutes = I2Rt
Let R be the resistance of the bulb, then:
(a) Energy provided by battery in 10 mins = Power x time = (VIt)= 4×0.5×20= 40 watt-hour
(b) Heat dissipated in the bulb in 10 minutes = I2Rt
Let R be the resistance of the bulb, then:
31. (a) What is meant by a ring circuit in house wiring?
(b) What are the advantages of a parallel connection?
Answer
(a) The ring system consists of a ring of three wires namely live wire, earth wire and neutral wire, which originate from the main fuse box and after running around the rooms in the house comes back to the main fuse box, thus, completing a ring. In ring system, a separate connection is taken from the live wire of the ring for each appliance. In the ring circuit, all appliances are connected in parallel.
(b) Advantages of a parallel connection are:
(b) Advantages of a parallel connection are:
- In parallel arrangement, each appliance works at the same voltage. For example, if several bulbs are connected in parallel, each bulb glows at the same voltage. Therefore, the glow of a bulb is unaffected if another bulb is switched on or off.
- In parallel arrangement, if one bulb (or appliance
32. (a) Explain with the help of a diagram how does 'short circuiting' occur in an electric kettle.
(b) Out of the wires- live, neutral and earthing, on which line will you connect the fuse and the switch. Give reason for your answer.
Answer
(b) Fuse and switches are always connected to the live wire so that when the switch is in off position no current flows through the appliance and its live and neutral wires are at same potential. In case, excess current flows through the live wire, the fuse wire melts and breaks the circuit; so no excess current flows through the appliance and it is protected against any damage.
33. What is meant by the term 'electromotive force' of a cell?
Answer
When no current is drawn from the cell i.e. when the cell is in open circuit, the potential difference between the terminals of the cell is called its electromotive force (or e.m.f).
34. Explain, why is the p.d. between the terminals of a storage battery less when it is supplying current than when it is on open circuit. A battery of e.m.f. 10 volts and internal resistance 2.5 ohms has two resistances of 50 ohms each connected to it. Calculate the power dissipated in each resistance
(a) When they are in series,
(b) When they are in parallel.
In each case calculate the power dissipated in the battery.
(b) When they are in parallel.
In each case calculate the power dissipated in the battery.
Answer
In an open circuit no current is drawn from the cell whereas in closed circuit, an amount of energy is spent in the flow of unit positive charge through the electrolyte of the cell. Thus, the p.d. between the terminals of a storage battery is less when it is supplying current.
Given, emf e = 10 V, internal resistance r = 2.5 Ω
Given, emf e = 10 V, internal resistance r = 2.5 Ω
35. Electrical appliances have voltage and power ratings as listed below. Which has the larger working resistance?
|
Appliances |
Voltage (V) |
Power (W) |
|
(a) Washing machine |
250 |
3000 |
Answer
|
Appliances |
Voltage (V) |
Power (W) |
Resistance (R in 0 =V2/P) |
|
(a) washing machine |
250 |
3000 |
20.8 |
Therefore, hair curler has the largest working resistance.
36. If the potential differences between two parts of a thundercloud is 108 V, what is the amount of energy given up during the passage of 20 coulombs ?
(a) 2× 10-7 J
(b) 200 J
(c) 5 × 106 J
(d) 2 × 109 J
(e) 2 × 1010 J
(a) 2× 10-7 J
(b) 200 J
(c) 5 × 106 J
(d) 2 × 109 J
(e) 2 × 1010 J
Answer
Given, p.d. V = 108 volt, charge Q = 20 coulombs
Energy given = QV = 20 × 108 = 2×108 J
∴ (d) 2 × 109 J
Energy given = QV = 20 × 108 = 2×108 J
∴ (d) 2 × 109 J
37. A 3 pin mains plug is fitted to the lead for a 1 kW electric kettle to be used on a 250 V A.C. supply. Which of the following statements is NOT correct?
(a) A 13 A fuse is the most appropriate value to use.
(b) The brown wire should be connected to the live side of the mains.
(b) The brown wire should be connected to the live side of the mains.
Answer
Incorrect statement:
(a) A 13A fuse is the most appropriate value to use
(a) A 13A fuse is the most appropriate value to use
38. (i) Name the principle on which a transformer works.
(ii) What is the function of a step-up transformer?
(iii) Can a transformer work when it is connected to a D.C. source? Give a reason.
(iv) Draw a simple labeled diagram of a step-down transformer.
(v) Draw a simple labeled diagram of a step-up transformer.
(iii) Can a transformer work when it is connected to a D.C. source? Give a reason.
(iv) Draw a simple labeled diagram of a step-down transformer.
(v) Draw a simple labeled diagram of a step-up transformer.
Answer
(i) A transformer works on the principle of electromagnetic induction.
(ii) Function of a step-up transformer is to increase the a.c. voltage and decrease the current.
(iii) No, a transformer cannot work on a d.c. source. With a d.c. source, there will be no change in magnetic flux linked with the secondary coil.
(ii) Function of a step-up transformer is to increase the a.c. voltage and decrease the current.
(iii) No, a transformer cannot work on a d.c. source. With a d.c. source, there will be no change in magnetic flux linked with the secondary coil.
39. The diagram 10 shows two coils X and Y. The coil X is connected to a battery S and a key K. The coil Y is connected to a galvanometer G.
When the key K is closed. State the polarity
(i)At the end of the coil X,
(ii)At the end C of the coil Y,
(iii)At the end C of the coil Y if the coil Y is (a) Moved towards the coil X, (b) Moved away from the coil X.
(i)At the end of the coil X,
(ii)At the end C of the coil Y,
(iii)At the end C of the coil Y if the coil Y is (a) Moved towards the coil X, (b) Moved away from the coil X.
Answer
(i) Current at the end B of the coil X is anticlockwise therefore at this end there is north pole.
(ii) While closing the key, polarity at the end C of the coil Y will be north. There will be no polarity at the end C of the coil Y when the current becomes steady in the coil X.
(iii) (a) While the coil Y is moved towards the coil X, the polarity at the end C of the coil Y is north.
(b) While the coil Y is moved away the coil X, the polarity at the end C of the coil Y is south.
(ii) While closing the key, polarity at the end C of the coil Y will be north. There will be no polarity at the end C of the coil Y when the current becomes steady in the coil X.
(iii) (a) While the coil Y is moved towards the coil X, the polarity at the end C of the coil Y is north.
(b) While the coil Y is moved away the coil X, the polarity at the end C of the coil Y is south.
40.The following diagram shows a fixed coil of several turns connected to a center zero galvanometer G and a magnet NS which can move.
(a) Describe the observation in the galvanometer if
(i) The magnet is moved rapidly in the direction of arrow,
(ii) The magnet is kept still after it has moved into the coil,
(iii) The magnet then rapidly pulled out the coil.
(a) Describe the observation in the galvanometer if
(i) The magnet is moved rapidly in the direction of arrow,
(ii) The magnet is kept still after it has moved into the coil,
(iii) The magnet then rapidly pulled out the coil.
(b) How would the observation alter if a more powerful magnet is used?
Answer
(a)
- When the magnet is moved rapidly in the direction of arrow, the magnetic flux linked with the coil changes and there is a deflection in the galvanometer, indicating a flow of current through the coil.
- On keeping the magnet still, the magnetic flux linked with the coil does not change and there is no deflection in the galvanometer, indicating that no current is flowing through the coil.
- When the magnet is rapidly pulled out, there is again change in the magnetic flux linked with the coil and the galvanometer shows a deflection but this time in opposite direction, indicating that a current is flowing in opposite direction in the coil.
41. Tick (√) the correct choice among the following: (a) A strong short bar magnet is dropped into a solenoid connected to a galvanometer as shown in fig. 12. The galvanometer shows a deflection to the left when the current in the coil as seen from above is counterclockwise. Which one of the following shows the correct deflection of the galvanometer when the magnet is at positions A, B and C.
|
At A |
At B |
At C |
|
(i) left |
Left |
Left |
|
(ii) left |
Zero |
Zero |
|
(iii) left |
Zero |
Right |
|
(iv) left |
Right |
Left |
|
(v) left |
Right |
right |
Answer
(a) (v) left left left
(b) Consider two cases of two parallel current carrying conductors. Current in the same direction and currents in the opposite directions will produce
(i) Attraction and repulsion respectively,
(ii) Repulsion and attraction respectively,
(iii) Attraction in both cases,
(iv) Repulsion in both cases,
(v) Oscillation in both cases.
(ii) Repulsion and attraction respectively,
(iii) Attraction in both cases,
(iv) Repulsion in both cases,
(v) Oscillation in both cases.
Answer
(ii) repulsion and attraction respectively
(ii) repulsion and attraction respectively
(c) Which of the following changes would enable a millimeter to be used as a direct current ammeter reading upto 1 A?
Answer
(i) connecting a large resistor in series
(d) In fig. 13, XY is at right angles to the magnetic field. The direction in which XY should be moved to induce an electron flow in the direction Y to X as indicated by the arrow is
(i) Along the direction of XY,
(ii) Towards N and perpendicular to XY,
(iii) Towards 5 and perpendicular to XY,
(iv) Upwards and perpendicular to XY.
(ii) Towards N and perpendicular to XY,
(iii) Towards 5 and perpendicular to XY,
(iv) Upwards and perpendicular to XY.
Answer
(iv) upwards and perpendicular to XY
(e) There are 500 turns and 2000 turns in the primary and secondary coil of a transformer respectively. If the output voltage is 1000 V, how large is the input voltage?
(i) 250 V
(ii) 500 V
(iii) 1000 V
(iv) 2000 V
(v) 4000 V
Answer
(i) 250 V
42. Fig. 14 shows the essential features of a battery operated bell. The hammer strikes the gong when the switch is closed. State and explain the effect of using the following material successively to form the core.
(a) Plastic
(b) Steel
(c) Copper
Answer
(a) Plastic being non-magnetic cannot be used as a material for core. It shall not intensify the formed magnetic field.
(b) Steel has high retentivity. Hence, after prolonged use even when the switch is off, it may retain some magnetic property and attract the armature.
(c) Using copper as a material for core will introduce eddy currents in the core and thus, interfere with the working of the bell.
(b) Steel has high retentivity. Hence, after prolonged use even when the switch is off, it may retain some magnetic property and attract the armature.
(c) Using copper as a material for core will introduce eddy currents in the core and thus, interfere with the working of the bell.
43.A wire is passed through a piece of cardboard as shown in fig. 15. An electric current flows vertically down the wire and then flows back up. The two sections of the write are observed to repel each other. (a) With the aid of a diagram, describe how you would plot the magnetic field lines on the card.
(b) Sketch the magnetic field patterns around both the sections of the wire. Use your sketch to explain why the two sections repel each other.
(c) The repulsive force between the two sections is small. What changes could be made to increase it?
(c) The repulsive force between the two sections is small. What changes could be made to increase it?
Answer
(b) Since the magnetic field lines pass through the loop in the same direction, the two sections repel each other.
(c) To increase the repulsive force, the current through the loop should be increased.
44. Draw and label the diagram of a simple D.C. motor.
(a) Explain the rotation of the coil, giving a reason for your answer.
(b) How can you reverse the direction of rotation of the armature?
(c) How can you increase the speed of rotation of the motor?
(b) How can you reverse the direction of rotation of the armature?
(c) How can you increase the speed of rotation of the motor?
Answer
(a) The coil rotates because equal and opposite forces act on its arms which from a couple.
(b) The direction of rotation of armature can be reversed by interchanging the connections at the terminals of the battery joined tot he brushes of the motor.
(c) The speed of rotation of the motor can be increased by increasing the current through the coil or by increasing the number of turns in the coil and by increasing the strength of the magnetic field.
45. The diagram shows a coil connected to a center zero galvanometer G. The galvanometer shows a deflection to the right when the north pole N of a powerful magnet is moved to the right as shown.
(i) Explain, why the defelction occurs in the galvanometer.
(ii) State whether the current in the coil is clockwise or anticlockwise when viewed from the end P.
(iii) State the observation in G when the coil is moved away from north pole N of the magnet keeping the magnet stationary.
(iv) State the observation in G when both the coil and the magnet are moved to right at the same speed.
(ii) State whether the current in the coil is clockwise or anticlockwise when viewed from the end P.
(iii) State the observation in G when the coil is moved away from north pole N of the magnet keeping the magnet stationary.
(iv) State the observation in G when both the coil and the magnet are moved to right at the same speed.
Answer
(i) This is due to change in magnetic flux in the coil. Due to change in magnetic flux an induced emf is produced in the coil. Hence, a current flows through the galvanometer.
(ii) The current appears anticlockwise when viewed from end A because end A will form north-pole.
(iii) The galvanometer now deflects towards left.
(iv) No deflection is observed as there is no relative motion between the magnet and the coil.
(ii) The current appears anticlockwise when viewed from end A because end A will form north-pole.
(iii) The galvanometer now deflects towards left.
(iv) No deflection is observed as there is no relative motion between the magnet and the coil.
46. A primary of 800 turns is connected to a 220 V A.C. supply and the secondary has 8 turns. What will be the output voltage?
Answer
47. How can you convert a moving coil galvanometer into ammeter and voltmeter?
Answer
A moving coil galvanometer can be converted into an ammeter by connecting a low resistance (called a shunt) in parallel to the galvanometer.
A moving coil galvanometer can be converted into a voltmeter by connecting a high resistance in series with the galvanometer.
A moving coil galvanometer can be converted into a voltmeter by connecting a high resistance in series with the galvanometer.
48. Why is an ammeter always connected in series while the voltmeter always in parallel to a circuit across which voltage is to be measured?
Answer
An ammeter is a low resistance device; hence it is connected in series.
A voltmeter is a high resistance device; hence it is connected in parallel.
A voltmeter is a high resistance device; hence it is connected in parallel.
49. Describe two experiments to illustrate the production of electric current by electromagnetic induction. State the laws which determine the direction and magnitude of the induced current.
Answer
Experiment 1
- Take a coil of wire AB having a large number of turns.
- Connect the ends of the coil to a galvanometer as shown in figure.
- Take a strong bar magnet and move its north pole towards the end B of the coil.
- There is a momentary deflection in the needle of the galvanometer, say to the right. This indicates the presence of a current in the coil AB. The deflection becomes zero the moment the motion of the magnet stops.
- Now withdraw the north pole of the magnet away from the coil. Now the galvanometer is deflected toward the left, showing that the current is now set up in the direction opposite to the first.
- Place the magnet stationary at a point near to the coil, keeping its north pole towards the end B of the coil. We see that the galvanometer needle deflects toward the right when the coil is moved towards the north pole of the magnet. Similarly the needle moves toward left when the coil is moved away.
- When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer drops to zero.
Experiment 2
- Take two different coils of copper wire having large number of turns (say 50 and 100 turns respectively). Insert them over a non-conducting cylindrical roll, as shown in figure.
- Connect the coil-1, having larger number of turns, in series with a battery and a plug key. Also connect the other coil-2 with a galvanometer as shown.
- Plug in the key. Observe the galvanometer. We will observe that the needle of the galvanometer instantly jumps to one side and just as quickly returns to zero, indicating a momentary current in coil-2.
- Disconnect coil-1 from the battery. We will observe that the needle momentarily moves, but to the opposite side. It means that now the current flows in the opposite direction in coil-2.
From these observations, we conclude that a potential difference is induced in the coil-2 whenever the electric current through the coil-1 is changing (starting or stopping). This process, by which a changing magnetic field in a conductor induces a current in another conductor, is called electromagnetic induction.
To determine the direction of induced current, we use Fleming's left hand rule: Stretch the thumb, forefinger and middle finger of right hand so that they are perpendicular to each other. If the forefinger indicates the direction of the magnetic field and the thumb shows the direction of motion of conductor, then the middle finger will show the direction of induced current.
Magnitude of induced current can be measured with the help of a galvanometer.
50. The circuit diagram in fig. 17 shows two coils of an insulated copper wire wound on a cardboard tube. G is a center zero galvanometer.
(a) Describe what will happen when the switch K is closed for several seconds and then opened again.
(b) What will be the effect of repeating the experiment with an iron placed in the tube?
Answer
(a) We will observe that the needle of the galvanometer instantly jumps to one side and just as quickly returns to zero, indicating a momentary current in coil connected to galvanometer.
(b) Introducing an iron bar in the tube, will increase the amount of induced current and the galvanometer will show a greater deflection.
(b) Introducing an iron bar in the tube, will increase the amount of induced current and the galvanometer will show a greater deflection.
51. Fig. 18 shows the core of a transformer and its input and output connections.
State the material used for the core and describe the structure of the core.
(a) Use the values given in the diagram to calculate the turns ratio for the transformer and connections, labeling all parts of the diagram.
(b) If a current of 2 A is taken from the output, calculate the current in the input circuit (Assume the transformer to be ideal).
(a) Use the values given in the diagram to calculate the turns ratio for the transformer and connections, labeling all parts of the diagram.
(b) If a current of 2 A is taken from the output, calculate the current in the input circuit (Assume the transformer to be ideal).
Answer
(a) Soft iron is used as the material for the core of transformer. Structure of core: The core is made by taking thin rectangular sheets or laminas of soft iron (or silicon-steel) and dipping them in an insulating paint or varnish. These insulated sheets are stacked together to form a solid looking rectangular frame. This is called the laminated core of the transformer.
52. Fig. 19 shows a simple form of an A.C. generator.
(a) Name the parts labeled A and B.
(b) What would be the effect of doubling the number of turns on the coil if the speed of rotation remains unchanged?
(c) Which of the output terminals is positive if the coil is rotating in the direction shown in the diagram (anticlockwise)?
(d) What is the position of the rotating coil when p.d. across its ends is zero? Explain why p.d. is zero when the coil is at this position.
(e) Sketch a graph showing how the p.d. across the ends of the rotating coil vanes with time for an A.C. dynamo.
(f) On the same sheet of paper and vertically below the first graph using the same time scale, sketch graphs to show the effect of
(i) Doubling the speed of rotation and at the same time keeping the field and the number of turns constant,
(ii) Doubling the number of turns on the coil and at the same time doubling the speed of rotation of the coil, keeping the speed constant.
(b) What would be the effect of doubling the number of turns on the coil if the speed of rotation remains unchanged?
(c) Which of the output terminals is positive if the coil is rotating in the direction shown in the diagram (anticlockwise)?
(d) What is the position of the rotating coil when p.d. across its ends is zero? Explain why p.d. is zero when the coil is at this position.
(e) Sketch a graph showing how the p.d. across the ends of the rotating coil vanes with time for an A.C. dynamo.
(f) On the same sheet of paper and vertically below the first graph using the same time scale, sketch graphs to show the effect of
(i) Doubling the speed of rotation and at the same time keeping the field and the number of turns constant,
(ii) Doubling the number of turns on the coil and at the same time doubling the speed of rotation of the coil, keeping the speed constant.
Answer
(b) Increasing the number of turns will increase the current through the coil.
(c) Terminal X will be positive.
(d) When the plane of the coil is normal to the magnetic field, the magnetic flux linked with the coil is maximum and the p.d. across its ends is zero.
53. Name the type of transformer shown in fig.20. What changes will you observe regarding the brightness of the lamp, if
(a) The no. of turns in the secondary is doubled?
(b) The part X of the transformer is removed?
(c) The core of the transformer is made out of copper?
(d) The A.C. supply is replaced by D.C.?
(a) The no. of turns in the secondary is doubled?
(b) The part X of the transformer is removed?
(c) The core of the transformer is made out of copper?
(d) The A.C. supply is replaced by D.C.?
Answer
It is a diagram of step-down transformer.
(a) Brightness of bulb will increase because increasing the number of turns in the secondary will increase the change in magnetic flux linked with the coil.
(b) Part X is the core of the transformer is removed, it shall become an open core and there shall be magnetic flux link loss; i.e. the entire magnetic field lines produced by the primary shall not be linked with the secondary.
(c) If the core of the transformer is made of copper due to the formation of eddy currents a lot of energy shall be lost.
(d) A transformer cannot be used with direct current (d.c.) since its working is based on the principle that when there is a change of magnetic field lines due to varying current of same in one coil, an induced varying current of same frequency flows in the other coil. If the current in one coil is constant (i.e. d.c.), no induced current will flow in the other coil since there will be no change in the magnetic field lines linked with the coil.
(a) Brightness of bulb will increase because increasing the number of turns in the secondary will increase the change in magnetic flux linked with the coil.
(b) Part X is the core of the transformer is removed, it shall become an open core and there shall be magnetic flux link loss; i.e. the entire magnetic field lines produced by the primary shall not be linked with the secondary.
(c) If the core of the transformer is made of copper due to the formation of eddy currents a lot of energy shall be lost.
(d) A transformer cannot be used with direct current (d.c.) since its working is based on the principle that when there is a change of magnetic field lines due to varying current of same in one coil, an induced varying current of same frequency flows in the other coil. If the current in one coil is constant (i.e. d.c.), no induced current will flow in the other coil since there will be no change in the magnetic field lines linked with the coil.
54. A transformer has 400 turns in the primary winding and 10 turns in the secondary winding. The primary e.m.f. is 250 V and the primary current is 2.0 A. calculate:
(a) The secondary voltage,
(b) The secondary current, assuming 100% efficiency.
(a) The secondary voltage,
(b) The secondary current, assuming 100% efficiency.
Answer
Since, the number of turns in the primary is more than the no. of turn in the secondary it is a step - down transformer. 
55. Transformers are usually designed so that their efficiency is as close to 100% as possible. Describe two features in transformer design which help to achieve high efficiency?
Answer
Features which provide greater efficiency to a transformer are:
- The core of the transformer is laminated which prevents the formation of eddy currents.
- A closed soft-iron core is used which reduces the magnetic field link loss and hysteresis loss.
56. Fill in the blanks by writing (i) Only soft iron, (ii) Only steel, (iii) Both soft-iron and steel for the material of core and/or magnet.
(a) Electric bell ______.
(b) Electromagnet ______.
(c) D.C. motor ______.
(d) A. C. generator______.
(e) Transformer______.
(a) Electric bell ______.
(b) Electromagnet ______.
(c) D.C. motor ______.
(d) A. C. generator______.
(e) Transformer______.
Answer
(a) magnet - soft iron
(b) core- soft iron
(c) core- soft iron, magnet - steel
(d) core- soft iron, magnet - steel
(e) core-soft iron
(b) core- soft iron
(c) core- soft iron, magnet - steel
(d) core- soft iron, magnet - steel
(e) core-soft iron
57. The circuit diagram (Fig. 21) shows three resistors 20, 4 s2 and RQ connected to a battery of e.m.f. 2 V and internal resistance 3 12. A main current of 0.25 A flows through the circuit.
(i) What is the p.d. across the 4 r2 resistor? (ii) Calculate the p.d. across the internal resistance of the cell. (iii) What is the p.d. across the R 0 or 2 f2resistors? (iv) Calculate the value of R.
Answer
(i) Potential difference across the 40 resistor = current through the resistor × its resistance = 0.25× 4= 1V
(ii) Potential difference across interns resistance current through internal resistance × internal resistance = 0.25×3 = 0.75V
(iii) P.d. across the R Ω or 3Ω resistor = emf -(p.d. across 4Ω resister + p.d. across internal resistance) = 2 -( 1+ 0.75) = 0.25V 
58. A cell supplies a current of 1.2 A through two 2Ω resistors connected in parallel. When the resistors are connected in series, if supplies a current of 0.4 A. Calculate the internal resistance and e.m.f of the cell.
Answer
Let e be the emf of the cell and r be its internal resistance.
Case I: In parallel,
total resistance, RP = (r +1) Ω
total current = 1.2A
∴ e = 1.2 (r +1) ...(i)
Case II: In series,
total resistance, RS = (r + 4)Ω
total current = 0.4A
e = 0.4 (r + 4) ...(ii)
From (i) and (ii)
1.2 (r +1) = 0.4 (r + 4)
or, 3 (r +1) = (r + 4)
or, 3r + 3 = r + 4
or, r =0.5Ω
Putting this value of r in (ii) we get:
e = 0.4( 0.5 + 4)
or, e = 1.8 V
∴ r = 0.5 Ω and e = 1.8 V
Case I: In parallel,
total resistance, RP = (r +1) Ω
total current = 1.2A
∴ e = 1.2 (r +1) ...(i)
Case II: In series,
total resistance, RS = (r + 4)Ω
total current = 0.4A
e = 0.4 (r + 4) ...(ii)
From (i) and (ii)
1.2 (r +1) = 0.4 (r + 4)
or, 3 (r +1) = (r + 4)
or, 3r + 3 = r + 4
or, r =0.5Ω
Putting this value of r in (ii) we get:
e = 0.4( 0.5 + 4)
or, e = 1.8 V
∴ r = 0.5 Ω and e = 1.8 V
59. Complete the following diagram of a transformer and name the parts labeled A and B. Name the part you have drawn to complete the diagram. What is the material of this part? In this transformer a step - up or step- down ? Why ?
Answer
The part drawn to complete the diagram the core.
Material of core is soft- iron.
It is a step down transformer because the number of turns in the primary is more than the number of turns in the secondary.
60. Why do you mean by turns ratio of a transformer? Can it work with D.C.?
Answer
The ratio of number of turns NS in secondary coil to the number of turns NP in the primary coil (NS / NP) is called the turns ratio.
A transformer cannot be used with direct current (d.c.) since its working is based on the principle that when there is a change of magnetic field lines due to varying current of same in one coil, an induced varying current of same frequency flows in the other coil. If the current in one coil is constant (i.e. d.c.), no induced current will flow in the other coil since there will be no change in the magnetic field lines linked with the coil.
A transformer cannot be used with direct current (d.c.) since its working is based on the principle that when there is a change of magnetic field lines due to varying current of same in one coil, an induced varying current of same frequency flows in the other coil. If the current in one coil is constant (i.e. d.c.), no induced current will flow in the other coil since there will be no change in the magnetic field lines linked with the coil.