ICSE Solutions for Chapter 8 Current Electricity Class 10 Selina Physics
Exercise 8 A
Solution 1: The resistivity of a material is the resistance of a wire of that material of unit length and unit area of cross-section.
Its S.I. unit is ohm metre.
Question 2: Write an expression connecting the resistance and resistivity. State the meaning of symbols used.
Solution 2: Expression :
R = p I/A
p - resistivity
R- resistance
L - length of conductor
A - area of cross-csection
Question 3: State the order of resistivity of (i) a metal, (ii) a semiconductor and (iii) an insulator.
Solution 3: Metal < Semiconductor < Insulator
Question 4: Name a substance of which the resistance remains almost unchanged by the increase in temperature.
Solution 4: Manganin
Question 5: Name the material used for making the connection wires. Give a reason for your answer. Why should a connection wire be thick?
Solution 5: 'Copper or Aluminium' is used as a material for making connection wires because the resistivity of these materials is very small, and thus, wires made of these materials possess negligible resistance.
The connection wires are made thick so that their resistance can be considered as negligible.
R = p I/a
Therefore, greater the area of cross-section, lesser shall be the resistance.
Question 6: Name a material which is used for making the standard resistor. Give a reason for your answer.
Solution 6: Manganin is used for making the standard resistor because its resistivity is quite large and the effect of change in temperature on their resistance is negligible.
Question 7: Name the material used for making a fuse wire. Give a reason.
Solution 7: Generally fuse wire is made of an alloy of lead and tin because its resistivity is high and melting point is low.
Question 8: Name the material used for (i) filament of an electric bulb, (ii) heating element of a room heater.
Solution 8: (i) A wire made of tungsten is used for filament of electric bulb because it has a high melting point and high resistivity.
(ii) A nichrome wire is used as a heating element for a room heater because the resistivity of nichrome is high and increase in its value with increase in temperature is high.
Question 9: What is a superconductor? Give one example of it.
Solution 9: A superconductor is a substance of zero resistance at a very low temperature. Example: Mercury at 4.2 K.
Question 10: A substance has zero resistance below 1 k. what is such a substance called?
Solution 10: Superconductor
Multiple Choice Type
Question 1: Which of the following is an ohmic resistance?
(a) diode valve (b) junction diode (c) filament of a bulb (d) nichrome
Solution 1: Nichrome is an ohmic resistance.
Hint: Substances that obey Ohm's law are called Ohmic resistors.
Question 2: For which of the following substance, resistance decreases with increase in temperature? (a) copper (b) mercury (c) carbon (d) platinum
Solution 2: For carbon, resistance decreases with increase in temperature.
Hint: For semiconductors such as carbon and silicon, the resistance and resistivity decreases with the increase in temperature.
Numericals
Question 1: In a conductor 6.25 x 1016 electrons flow from its end A to B in 2 s. Find the current flowing through the conductor (e = 1.6 x 1019 C)
Solution 1: Number of electrons flowing through the conductor,
N = 6.25 × 1016 electrons
Time taken, t = 2 s
Given, e = 1.6 × 10-19c
Let I be the current flowing through the conductor.
Then, I = ne/t
∴ I = (6.25 × 1016)(1.6 × 10-19) = 5 × 10-J A
Or, I = 5 mA
Thus, 5 Ma current flows from B to A.
Question 2: A current of 1.6 mA flows through a conductor. If charge on an electron is -1.6×10-19 coulomb, find the number of electrons that will pass each second through the cross section of that conductor.
Solution 2: Current , I = 1.6 mA = 1.6× 10-3A
Charge, Q = -1.6 × 10-19 coulomb
t = 1 sec
I = Q/t
Q = I × t
Q = 1.6 × 10-3 × 1
No. of electrons = (1.6 × 10-3)/(1.6 × 10-19)
= 1016
Question 3: Find the potential difference required to pass a current of 0.2 A in a wire of resistance 20Ω
Solution 3: Current (I) = 0.2 A
Resistance (R) = 20 ohm
Potential Difference (V) = ?
According to Ohm's Law :
V = IR
V = 0.2 × 20 = 4 V
Question 4: An electric bulb draws 1.2 A current at 6.0 V. Find the resistance of filament of bulb while glowing.
Solution 4: Current (I) = 1.2 A
Potential Difference/Voltage (V) = 6.0 V
Resistance (R) = ?
According to Ohm's Law :
V = IR
Then R = V/I
R = 6/1.2
R = 5 Ohm
Question 5: A car bulb connected to a 12 volt battery draws 2 A current when glowing. What is the resistance of the filament of the bulb? Will the resistance be more same or less when the bulb is not glowing?
Solution 5: Potential Difference/Voltage (V) = 12 V
Current (I) = 2 A
Resistance (R) = ?
According to Ohm's Law :
V=IR
Then R = V/I
R = 12/2
R = 6 Ohm
Resistance will be less when the bulb is not glowing.
Question 6: Calculate the current flowing through a wire of resistance 5 Ω connected to a battery of potential difference 3 V.
Solution 6: Potential Difference/Voltage (V) = 3 V
Resistance (R) = 5 ohm
Current (I) = ?
According to Ohm's Law :
V = IR
Then I = V/R
I = 3/5 = 0.6 A
Question 7: In an experiment of verification of Ohm's law following observations are obtained.
(a) potential difference V when the current I is 0.5 A,
(b) current I when the potential difference V is 0.75 V,
(c) resistance in circuit.
Solution 7:
(b) 0.3 A
(c) The graph is linear so resistance can be found from any value of the given table. For instance:
When V = 2.5 Volt
Current is I = 1.0 amp
According to ohm's law :
R = V/I
R = 2.5/1.0 = 2.5 ohm
Question 8: Two wires of the same material and same length have radii r1 and r2 respectively compare: (i) their resistances, (ii) their resistivities.
Solution 8: (i) For wire of radius
R1 = ρl/A1
R1 = ρl/πr12
(ii) For wire of radius r2 :
R2 = ρl/A2
R2 = ρl/πr22
R2 : R2 will be ρl/πr12 : ρ l/πr22
r22 : r12
(ii) Since the material of the two wires is same, so their resistivities will also be same i.e., ρ1:ρ2 = 1:1
Question 9: A given wire of resistance 1 Ω is stretched to double its length. What will be its new resistance?
Solution 9: Let ‘l be the length and 'a' be the area of cross-section of the resistor with resistance, R = 1Ω
when the wire is stretched to double its length,
the new length I' = 2I and the new area of cross section,
a' = a/2
∴ Resistance (R') = ρ I’/a’ = ρ 2I/(a/2)
∴ R' = 4 ρ I/a = 4R
∴ R' = 4 × 1 = 4 Ω
Question 10: A wire 3 ohm resistance and 10 cm length is stretched to 30 cm length. Assuming that it has a uniform cross section, what will be its new resistance?
Solution 10: Resistance (R) = 3 ohm
Length I = 10 cm
New Length (l') = 30 cm = 3 × l
R’ = ρ l/A
New Resistance :
With stretching length will increase and area of cross-section will decrease in the same order
R’ = ρ 3/(A/3)
Therefore,
R' = 9 ρl/A = 9R
R' = 9 × 3 = 27 Ω
Question 11: A wire of 9 ohm resistance having 30 cm length is tripled on itself. What is its new resistance?
Solution 11: Resistance (R) = 9 ohm
Length l = 30 cm
New Length (I) = 30 cm = 3/l = 10 cm
R = ρl/A
New Resistance :
With change in length, there will be change in area of cross-section also in the same order.
R’ = ρ(I/3)/(3A)
R’ = l/9 ρ I/A
R’ = l/9R
R' = 1ohm
Question 12: What length of copper wire of resistivity 1.7 x 10-8 Ω m and radius 1 mm is required so that its resistance is 1 Ω?
Solution 12: Resistance (R) = 1 ohm
Resistivity (0) = 1.7×10-8 ohm metre
Radius (r) = 1 mm = 10-3 m
Length (I) = ?
R = ρ l/A
I = RA/ρ
= Rπr2/ρ
= (1 × π × 10-6)/(1.7 × 10-8)
= 184.7m
Question 9: A given wire of resistance 1 Ω is stretched to double its length. What will be its new resistance?
Solution 9: Let ‘l be the length and 'a' be the area of cross-section of the resistor with resistance, R = 1Ω
when the wire is stretched to double its length,
the new length I' = 2I and the new area of cross section,
a' = a/2
∴ Resistance (R') = ρ I’/a’ = ρ 2I/(a/2)
∴ R' = 4 ρ I/a = 4R
∴ R' = 4 × 1 = 4 Ω
Question 10: A wire 3 ohm resistance and 10 cm length is stretched to 30 cm length. Assuming that it has a uniform cross section, what will be its new resistance?
Solution 10: Resistance (R) = 3 ohm
Length I = 10 cm
New Length (l') = 30 cm = 3 × l
R’ = ρ l/A
New Resistance :
With stretching length will increase and area of cross-section will decrease in the same order
R’ = ρ 3/(A/3)
Therefore,
R' = 9 ρl/A = 9R
R' = 9 × 3 = 27 Ω
Question 11: A wire of 9 ohm resistance having 30 cm length is tripled on itself. What is its new resistance?
Solution 11: Resistance (R) = 9 ohm
Length l = 30 cm
New Length (I) = 30 cm = 3/l = 10 cm
R = ρl/A
New Resistance :
With change in length, there will be change in area of cross-section also in the same order.
R’ = ρ(I/3)/(3A)
R’ = l/9 ρ I/A
R’ = l/9R
R' = 1ohm
Question 12: What length of copper wire of resistivity 1.7 x 10-8 Ω m and radius 1 mm is required so that its resistance is 1 Ω?
Solution 12: Resistance (R) = 1 ohm
Resistivity (0) = 1.7×10-8 ohm metre
Radius (r) = 1 mm = 10-3 m
Length (I) = ?
R = ρ l/A
I = RA/ρ
= Rπr2/ρ
= (1 × π × 10-6)/(1.7 × 10-8)
= 184.7m
Exercise 8 B
Question 1: Explain the meaning of the terms e.m.f.., terminal voltage and internal resistance of a cell.
Solution 1: e.m.f.: When no current is drawn from a cell, the potential difference between the terminals of the cell is called its electro-motive force (or e.m.f.). Terminal voltage: When current is drawn from a cell, the potential difference between the electrodes of the cell is called its terminal voltage.
Internal Resistance: The resistance offered by the electrolyte inside the cell to the flow of electric current through it is called the internal resistance of the cell.
Question 2: State two differences between the e.m.f and terminal voltage of a cell.
Solution 2:
|
e.m.f. of cell |
Terminal voltage of cell |
|
It is measured by the amount of work done in moving unit positive charge in the complete circuit inside and outside of the cell. |
It is measured by the amount of work done in moving a unit positive charge in the circuit outside the cell. |
|
It is the characteristic of the cell i.e. it does not depend on the amount of current drawn from the cell. |
It depends on the amount of current drawn from the cell. More the current is drawn from the cell, less is the terminal voltage. |
|
It is equal to the terminal voltage when cell is not in use, while greater than the terminal voltage when cell is in use. |
It is equal to the emf of cell when cell is not in use while less than the emf when cell is in use. |
Solution 3: Internal resistance of a cell depends upon the following factors:
(i) The surface area of the electrodes: Larger the surface area of the electrodes, less is the internal resistance.
(ii) The distance between the electrodes: More the distance between the electrodes, greater is the internal resistance.
Question 4: A cell of e.m.f ɛ and internal resistance r is used to send current to an external resistance R. write expressions for (a) the total resistance of circuit, (b) the current drawn from the cell. (c) the p.d across the cell. And (d) voltage drop inside the cell.
Solution 4: (a) Total resistance = R + r
(b) Current drawn from the circuit:
As we know that,
ε = V+v
= IR + Ir
= I(R+r)
I = e/(R + r)
(c) p.d. across the cell : ɛ/(R + r) × R
(d) voltage drop inside the cell: ɛ/(R + r) × r
Question 5: A cell is used to send current to an external circuit. (a) How does the voltage across its terminals compare with its e.m.f? (b) under what condition is the e.m.f of a cell equal to its terminal voltage?
Solution 5: (a) Terminal voltage is less than the emf: Terminal Voltage < e.m.f.
(b) e.m.f. is equal to the terminal voltage when no current is drawn.
Question 6: Explain why the p.d across the terminals of a cell is more in an open circuit and reduced in a closed circuit.
Solution 6: When the electric cell is in a closed circuit the current flows through the circuit. There is a fall of potential across the internal resistance of the cell. So, the p.d. across the terminals in a closed circuit is less than the p.d. across the terminals in an open circuit by an amount equal to the potential drop across the internal resistance of the cell.
Question 7: Write the expressions for the equivalent resistance R of three resistors R1, R2 and R3 joined in (a) parallel (b) series
Solution 7: (a) Total Resistance in series:
R = R1 + R2 + R3
(b) Total Resistance in parallel:
1/R = 1/R1 + 1/R2 + 1/R3
Question 8: How would you connect two resistors in series? Draw a diagram. Calculate the total equivalent resistance.
Solution 8:
On applying Ohm’s law to be two resistor separately, we further have
V1 = IR1
V2 = IR2
V = V1 + V2
IR = IR1 + IR2
R = R1 + R2
Total Resistance in series R
R = R1 + R2 + R3
Question 9: Show by a diagram how two resistors R1 and R2 are joined in parallel. Obtain an expression for the total resistance of combination.
Solution 9:
On applying Ohm's law to the two resistors separately, we further
Have
I1 = V/R1
I2 = V/R2
I = I1 + I2
V/R = V/R1 + V/R2
1/R = 1/R1 + 1/R2
Question 10: State how are the two resistors joined with a battery in each of the following cases when:
(a) same current flows in each resistor
(b) potential difference is same across each resistor
(c) equivalent resistance is less than either of the two resistances
(d) equivalent resistance is more than either of the two resistances.
Solution 10: (a) series
(b) parallel
(c) parallel
(d) series
Question 11: The V-I graph for a series combination and for a parallel combination of two resistors is shown in Fig . Which of the two, A or B, represents the parallel combination? Give a reason for your answer.
Solution 11: For the same change in I, change in V is less for the straight line A than for the straight line B (i.e., the straight line A is less steeper than B), so the straight line A represents small resistance, while the straight line B represents more resistance. In parallel combination, the resistance decreases while in series combination, the resistance increases. So A represents the parallel combination.
Multiple Choice Type
Question 1: In series combination of resistances:
(a) p.d is same across each resistance
(b) total resistance is reduced
(c) current is same in each resistance
(d) all above are true
Solution 1: In series combination of resistances, current is same in each resistance.
Hint: In a series combination, the current has a single path for its flow. Hence, the same current passes through each resistor.
Question 2: In parallel combination of resistances:
(a) p.d is same across each resistance
(b) total resistance is increased
(c) current is same in each resistance
(d) all above are true
Solution 2: In parallel combination of resistances, P.D. is same across each resistance.
Hint: In parallel combination, the ends of each resistor are connected to the ends of the same source of potential. Thus, the potential difference across each resistance is same and is equal to the potential difference across the terminals of the source (or battery).
Question 3: Which of the following combinations have the same equivalent resistance between X and Y?
Solution 3: (a) and (d)
Solution:In fig (a), the resistors are connected in parallel
Between X and Y.
Let R be their equivalent resistance.
Then, 1/R = ½ + ½ = 2/2 Ω
Or RI = 1 Ω ….(i)
In fig (d) a series combination of two 1 Ω resistors
Is in parallel with another series combination of two
1 Ω resistors
Series resistance of two 1 Ohm resistors,
R = (1 + 1) Ω = 2 Ω
Thus, we can say that across X and Y, two 2 Ω resistors are connected in parallel
Let R' be the net resistance across X and Y.
Then, 1/R’ = ½ + ½ = 2/2 Ω
Or, R' = 1 Ω.......... ......(ii)
From (i) and (ii), it is clear that (a) and (d) have
The same equivalent resistance between X and Y.
Numericals
Question 1: The diagram below in Fig. shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when (i) the key K is open, (ii) the key K is closed.
Solution 1: (i) Ammeter reading = 0 because of no current
Voltage V = ε - Ir
V = 2 – 0×1 = 2 volt
(ii) Ammeter reading :
I = ε /(R+r)
I = 2/(4+1) = 2/5 = 0.4 amp
Voltage reading :
Voltage V = ε - Ir
V = 2 - 0.4 x 1 = 2 - 0.4 = 1.6 V
Question 2: A battery of e.m.f 3.0 V supplies current through a circuit in which the resistance can be changed. A high resistance voltmeter is connected across the battery. When the current is 1.5 A, the voltmeter reads 2.7 V. Find the internal resistance of the battery.
Solution 2: ε = 3 volt
I = 1.5 A
V = 2.7 V
V = ε - Ir
r = (e-V)/I
= (3 – 2.7)/1.5 = 0.2 ohm
Question 3: A cell of e.m.f. 1.8V and internal resistance 2 Ω is connected in series with an ammeter of resistance 0.7 Ω and a resistor of 4.5 Ω as shown in Fig.
(b) what is the potential difference across the terminals of the cell?
Solution 3: (a) ε = 1.8 V
Total Resistance = 2 + 4.5 + 0.7 = 7.2 W
I = ?
I = ɛ/ R (total resistance)
I = 1.8/7.2 = 0.25 A
(b) Current (calculated in (a) part) I = 0.25 A
Now, total resistance excluding internal resistance = 4.5 + 0.7 = 5.2 ohm
V = IR = 0.25 × 5.2 = 1.3 V
Question 4: A battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm is series Find:
(a) the current through the battery
(b) the p.d. between the terminals of the battery.
Solution 4: (a) ε = 15 V
R = 6 + 3 = 9 ohm
R = 3 ohm
I = ?
I = ε/(R + r)
I = 15/(9 + 3) = 15/12 = 1.25 A
(b) Current (calculated in (a) part) I = 1.25 A
External Resistance R = 6 + 3 = 9 ohm
V = IR = 1.25 x 9 = 11.25 V
Question 5: A cell of e.m.f. ε and internal resistance r sends current 1.0 A when it is connected to an external resistance 1.9 Ω. But it sends current 0.5 A when it is connected to an external resistance 3.9 Ω. Calculate the values of ε and r.
Solution 5: In first case
I = 1 A, R = 1.9 ohm
ε = I(R + r) = 1(1.9 + r)
ε = 1.95 + r ----(1)
In second case
I = 0.5 A, R = 3.9 ohm
ε = I(R + r) = 0.5 (3.9 + r)
ε = 1.95 + 0.5r ----(2)
From eq. (1) and (2),
1.9 + r = 1.95 + 0.5r
r = 0.05/0.5 = 0.1 ohm
Substituting value of r
ε = 1.9 + r = 1.9 + 0.1 = 2 V
Question 6: Two resistors having resistance 4 Ω and 6 Ω are connected in parallel. Find their equivalent resistance.
Solution 6: Let R' be their equivalent resistance of the 4 Ω and 6 Ω resistors connected in parallel.
Then, 1/R’ = ¼ + 1/6 = (3+2)/12 = 5/12 Ω
Or, R' = 12/5 = 2.4 Ω
Question 7: Four resistors each of resistance 2 Ω are connected in parallel. What is the effective resistance?
Solution 7: R1 = 2 ohm
R2 = 2 ohm
R3 = 2 ohm
R4 = 2 ohm
1/R = 1/R1 + 1/R2 + 1/R3 + 1/R4 = ½ + ½ + ½ + ½ = 2
R = 0.5 ohm
Question 8: You have three resistors of values 2 Ω, 3 Ω and 5 Ω. How will you join them so that the total resistance is less than 1 Ω? Draw diagram and find the total resistance
Solution 8: The three resistors should be connected in parallel
To get a total resistance less than 1 Ω
Let R' be the total resistance.
Then,
1/R = ½ + 1/3 + 1/5 = (15+10+16)/30 = 31/30 Ω
Or, R = 30/31 = 0.97 Ω
Question 9: Three resistors each of 2 Ω are connected together so that their total resistance is 3 Ω. Draw a diagram to show this arrangement and check it by calculation.
Solution 9: A parallel combination of two resistors, in series with one resistor.
R1 = 2 ohm
R2 = 2 ohm
R3 = 2 ohm
1/R’ = 1/R1 + 1/R2
1/R’ = ½ + ½ = 1
1/R’ = 1 ohm
R = R’ + R3 = 1 + 2 = 3 ohm
Question 10: Calculate the equivalent resistance of the following combination of resistors r1, r2, r3 and r4 if r1 = r2 = r3 = r4 = 2.0 Ω, between the points A and B in Fig.
Solution 10: r1 = r2 = r3 = r4 = 2.0ohm
r' = r1 + r2 = 2 + 2 = 4 ohm
1/r” = 1/r3 + 1/r4 = ½ + ½ = 1
r” = 1ohm
r = r’ + r” = 4 + 1 = 5ohm
Question 11: A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is 2 ohm, find the resistance of the combination.
Solution 11: Resistance of each set:
r1 = 2 + 2 + 2 = 6 ohm
r2 = 2 + 2 + 2 = 6 ohm
r3 = 2 + 2 + 2 = 6 ohm
r4 = 2 + 2 + 2 = 6 ohm
Now these resistances are arranged in parallel :
1/r = 1/r1 + 1/r2 + 1/r3 + 1/r4
1/r = 1/6 + 1/6 + 1/6 + 1/6
r = 6/4 = 1.5ohm.
Question 12: In the circuit shown below in Fig, calculate the value of x if the equivalent resistance between A and B is 4 Ω.
Solution 12:
r1 = 4 ohm
r2 = 8 ohm
r3 = x ohm
r4 = 5 ohm
r = 4 ohm
r' = r1 + r2 = 4 + 8 = 12ohm
r" = r3 + r4 = (x + 5)ohm
1/r = 1/r + 1/r”
¼ = 1/12 + 1/(5 + x)
1/6 = 1/(5 + x)
x = 1 ohm
Question 13: Calculate the effective resistance between the points A and B in the circuit shown in Fig.
Solution 13:
In the figure above,
Resistance between XAY = (1 + 1 + 1) = 3 Ω
Resistance between XY = 2 Ω
Resistance between XBY = 6 Ω
Let R' be the net resistance between points X and Y
Then, 1/R’ = ½ + 1/3 + 1/6 = (3 + 2 + 1)/6 = 6/6 Ω
Or, RI = 1 Ω
Thus, we can say that between points A and B,
Three 1 Ω resistors are connected in series.
Let RAB be the net resistance between points A and B.
Then, RAB = (1 +1 + 1) Ω = 3 Ω
Question 14: A wire of uniform thickness with a resistance of 27Ω is cut into three equal pieces and they are joined in parallel. Find the equivalent resistance of the parallel combination.
Solution 14: Wire cut into three pieces means new resistance = 27/3 = 9
Now three resistance connected in parallel :
1/r = 1/r1 + 1/r2 + 1/r3 = 1/9 + 1/9 + 1/9
r = 9/3 = 3 ohm
Question 15: A circuit consists of a 1 ohm resistor in series with a parallel arrangement of 6 ohm and 3 ohm resistors. Calculate the total resistance if the circuit. Draw a diagram.
Solution 15: 1/r = 1/6 + 1/3 + 1/2
R = 2 oh m
R = 2 + 1 = 3 ohm
Question 16: Calculate the effective resistance between the points A and B in the network shown below in Fig.
Solution 16:
R1 = 1 + 2 = 3 ohm
R2 = 1.5 ohm
R1 and R2 are connected in parallel
1/R = 1/R1 + 1/R2 = 1/3 + 1/1.5 = 1
R = 1 ohm
Question 17: Calculate the equivalent resistance between A and B in the adjacent diagram in Fig.
Solution 17:
R1 = 3 + 2 = 5 ohm
R2 = 30 W
R3 = 6 + 4 = 10 ohm
R1, R2 and R3 are connected in parallel
1/R = 1/R1 + 1/R2 + 1/R3 = 1/5 + 1/30 + 1/10 = 10/30
R = 3 ohm
Question 18: In the network shown in adjacent Fig., calculate the equivalent resistance between the points.
(a) A and B
(b) A and C
Solution 18: (a) R1 = 2 + 2 + 2 = 6ohm
R2 = 2ohm
R1 and R2 are connected in parallel
1/R = 1/R1 + 1/R2 = 1/6 + ½ = 4/6
R = 6/4 = 1.5 ohm
(b) R1 = 2 + 2 = 4 ohm
R2 = 2 + 2 = 4 W
R1 and R2 are connected in parallel
1/R = 1/R1 + 1/R2 = ¼ + ¼ = 1/2
R = 2 ohm
Question 19: Five resistors, each 3 Ω, are connected as shown in Fig. Calculate the resistance (a) between the points P and Q. (b) between the points X and Y.
Solution 19: (a) R1 = 3 + 3 = 6 W
R2 = 3 W
R1 and R2 are connected in parallel
1/R = 1/R1 + 1/R2 = 1/6 + 1/3 = 1/2
(b) As calculated above R = 2 ohm
R3 = 3 ohm
R4 = 3 ohm
R' = R + R3 + R4 = 2 + 3 + 3 = 8 ohm
Question 20: Two resistors of 2.0 Ω and 3.0 Ω are connected (a) in series (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.
Solution 20: (a)
R1 = 2 ohm
R2 = 3 ohm
R = R1 + R2 = 2 + 3 = 5 ohm
V = 6V
I = V/R = 6/5 = 1.2 ohm
(b) R1 and R2 are connected in parallel
1/R = 1/R1 + 1/R2 = ½ + 1/3 = 5/6
R = 1.2 ohm
V = 6V
I = V/R = 6/1.2 = 5A
Question 21: A resistor of 6 Ω is connected in series with another resistor of 4 Ω. A potential difference of 20 V is applied across the combination. Calculate (a) the current in the circuit and (b) the potential difference across the 6 Ω resistor.
Solution 21: (a) R1 = 6 ohm
R2 = 4 ohm
R = R1 + R2 = 6 + 4 = 10 ohm
V = 20 V
I = V/R = 20/10 = 2 A
(b) R = 6 W
I = 2 A
V = ?
V = IR = 6 × 2 = 12 V
Question 22: In fig., calculate:
(a) the total resistance of the circuit
(b) the value if R, and
(c) the current flowing in R.
Solution 22: For resistor A:
R = 1 ohm
V = 2 V
I = V/R = 2/1 = 2A
For resistor B:
R = 2 ohm
V = 2 V
I = V/R = 2/2 = 1A
Question 23: A wire of length 5 m has a resistance of 2.0Ω calculate:
(a) the resistance of wire of length 1 m
(b) the equivalent resistance if two such wires each of length 2 m are joined in parallel.
(c) the resistance of 1 m length of wire of same material but of half diameter.
Solution 23: (a) V = 4 V
I = 0.4 A
Total Resistance R' = ?
R' = V/I = 0.4/4 = 10 ohm
(b) R1 = 20 ohm
R' = 10 ohm
1/R’ = 1/R + 1/R1
1/10 = 1/R + 1/20
1/R = 1/10 - 1/20 = 1/20
R = 20 Ω
Now, total resistance excluding internal resistance = 4.5 + 0.7 = 5.2 ohm
V = IR = 0.25 × 5.2 = 1.3 V
Question 4: A battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm is series Find:
(a) the current through the battery
(b) the p.d. between the terminals of the battery.
Solution 4: (a) ε = 15 V
R = 6 + 3 = 9 ohm
R = 3 ohm
I = ?
I = ε/(R + r)
I = 15/(9 + 3) = 15/12 = 1.25 A
(b) Current (calculated in (a) part) I = 1.25 A
External Resistance R = 6 + 3 = 9 ohm
V = IR = 1.25 x 9 = 11.25 V
Question 5: A cell of e.m.f. ε and internal resistance r sends current 1.0 A when it is connected to an external resistance 1.9 Ω. But it sends current 0.5 A when it is connected to an external resistance 3.9 Ω. Calculate the values of ε and r.
Solution 5: In first case
I = 1 A, R = 1.9 ohm
ε = I(R + r) = 1(1.9 + r)
ε = 1.95 + r ----(1)
In second case
I = 0.5 A, R = 3.9 ohm
ε = I(R + r) = 0.5 (3.9 + r)
ε = 1.95 + 0.5r ----(2)
From eq. (1) and (2),
1.9 + r = 1.95 + 0.5r
r = 0.05/0.5 = 0.1 ohm
Substituting value of r
ε = 1.9 + r = 1.9 + 0.1 = 2 V
Question 6: Two resistors having resistance 4 Ω and 6 Ω are connected in parallel. Find their equivalent resistance.
Solution 6: Let R' be their equivalent resistance of the 4 Ω and 6 Ω resistors connected in parallel.
Then, 1/R’ = ¼ + 1/6 = (3+2)/12 = 5/12 Ω
Or, R' = 12/5 = 2.4 Ω
Question 7: Four resistors each of resistance 2 Ω are connected in parallel. What is the effective resistance?
Solution 7: R1 = 2 ohm
R2 = 2 ohm
R3 = 2 ohm
R4 = 2 ohm
1/R = 1/R1 + 1/R2 + 1/R3 + 1/R4 = ½ + ½ + ½ + ½ = 2
R = 0.5 ohm
Question 8: You have three resistors of values 2 Ω, 3 Ω and 5 Ω. How will you join them so that the total resistance is less than 1 Ω? Draw diagram and find the total resistance
Solution 8: The three resistors should be connected in parallel
To get a total resistance less than 1 Ω
Then,
1/R = ½ + 1/3 + 1/5 = (15+10+16)/30 = 31/30 Ω
Or, R = 30/31 = 0.97 Ω
Question 9: Three resistors each of 2 Ω are connected together so that their total resistance is 3 Ω. Draw a diagram to show this arrangement and check it by calculation.
Solution 9: A parallel combination of two resistors, in series with one resistor.
R1 = 2 ohm
R2 = 2 ohm
R3 = 2 ohm
1/R’ = 1/R1 + 1/R2
1/R’ = ½ + ½ = 1
1/R’ = 1 ohm
R = R’ + R3 = 1 + 2 = 3 ohm
Question 10: Calculate the equivalent resistance of the following combination of resistors r1, r2, r3 and r4 if r1 = r2 = r3 = r4 = 2.0 Ω, between the points A and B in Fig.
Solution 10: r1 = r2 = r3 = r4 = 2.0ohm
r' = r1 + r2 = 2 + 2 = 4 ohm
1/r” = 1/r3 + 1/r4 = ½ + ½ = 1
r” = 1ohm
r = r’ + r” = 4 + 1 = 5ohm
Question 11: A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is 2 ohm, find the resistance of the combination.
Solution 11: Resistance of each set:
r1 = 2 + 2 + 2 = 6 ohm
r2 = 2 + 2 + 2 = 6 ohm
r3 = 2 + 2 + 2 = 6 ohm
r4 = 2 + 2 + 2 = 6 ohm
Now these resistances are arranged in parallel :
1/r = 1/r1 + 1/r2 + 1/r3 + 1/r4
1/r = 1/6 + 1/6 + 1/6 + 1/6
r = 6/4 = 1.5ohm.
Question 12: In the circuit shown below in Fig, calculate the value of x if the equivalent resistance between A and B is 4 Ω.
r1 = 4 ohm
r2 = 8 ohm
r3 = x ohm
r4 = 5 ohm
r = 4 ohm
r' = r1 + r2 = 4 + 8 = 12ohm
r" = r3 + r4 = (x + 5)ohm
1/r = 1/r + 1/r”
¼ = 1/12 + 1/(5 + x)
1/6 = 1/(5 + x)
x = 1 ohm
Question 13: Calculate the effective resistance between the points A and B in the circuit shown in Fig.
Solution 13:
In the figure above,
Resistance between XAY = (1 + 1 + 1) = 3 Ω
Resistance between XY = 2 Ω
Resistance between XBY = 6 Ω
Let R' be the net resistance between points X and Y
Then, 1/R’ = ½ + 1/3 + 1/6 = (3 + 2 + 1)/6 = 6/6 Ω
Or, RI = 1 Ω
Thus, we can say that between points A and B,
Three 1 Ω resistors are connected in series.
Let RAB be the net resistance between points A and B.
Then, RAB = (1 +1 + 1) Ω = 3 Ω
Question 14: A wire of uniform thickness with a resistance of 27Ω is cut into three equal pieces and they are joined in parallel. Find the equivalent resistance of the parallel combination.
Solution 14: Wire cut into three pieces means new resistance = 27/3 = 9
Now three resistance connected in parallel :
1/r = 1/r1 + 1/r2 + 1/r3 = 1/9 + 1/9 + 1/9
r = 9/3 = 3 ohm
Question 15: A circuit consists of a 1 ohm resistor in series with a parallel arrangement of 6 ohm and 3 ohm resistors. Calculate the total resistance if the circuit. Draw a diagram.
Solution 15: 1/r = 1/6 + 1/3 + 1/2
R = 2 oh m
R = 2 + 1 = 3 ohm
Question 16: Calculate the effective resistance between the points A and B in the network shown below in Fig.
Solution 16:
R1 = 1 + 2 = 3 ohm
R2 = 1.5 ohm
R1 and R2 are connected in parallel
1/R = 1/R1 + 1/R2 = 1/3 + 1/1.5 = 1
R = 1 ohm
Question 17: Calculate the equivalent resistance between A and B in the adjacent diagram in Fig.
Solution 17:
R1 = 3 + 2 = 5 ohm
R2 = 30 W
R3 = 6 + 4 = 10 ohm
R1, R2 and R3 are connected in parallel
1/R = 1/R1 + 1/R2 + 1/R3 = 1/5 + 1/30 + 1/10 = 10/30
R = 3 ohm
Question 18: In the network shown in adjacent Fig., calculate the equivalent resistance between the points.
(a) A and B
(b) A and C
Solution 18: (a) R1 = 2 + 2 + 2 = 6ohm
R2 = 2ohm
R1 and R2 are connected in parallel
1/R = 1/R1 + 1/R2 = 1/6 + ½ = 4/6
R = 6/4 = 1.5 ohm
(b) R1 = 2 + 2 = 4 ohm
R2 = 2 + 2 = 4 W
R1 and R2 are connected in parallel
1/R = 1/R1 + 1/R2 = ¼ + ¼ = 1/2
R = 2 ohm
Question 19: Five resistors, each 3 Ω, are connected as shown in Fig. Calculate the resistance (a) between the points P and Q. (b) between the points X and Y.
R2 = 3 W
R1 and R2 are connected in parallel
1/R = 1/R1 + 1/R2 = 1/6 + 1/3 = 1/2
(b) As calculated above R = 2 ohm
R3 = 3 ohm
R4 = 3 ohm
R' = R + R3 + R4 = 2 + 3 + 3 = 8 ohm
Question 20: Two resistors of 2.0 Ω and 3.0 Ω are connected (a) in series (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.
Solution 20: (a)
R2 = 3 ohm
R = R1 + R2 = 2 + 3 = 5 ohm
V = 6V
I = V/R = 6/5 = 1.2 ohm
1/R = 1/R1 + 1/R2 = ½ + 1/3 = 5/6
R = 1.2 ohm
V = 6V
I = V/R = 6/1.2 = 5A
Question 21: A resistor of 6 Ω is connected in series with another resistor of 4 Ω. A potential difference of 20 V is applied across the combination. Calculate (a) the current in the circuit and (b) the potential difference across the 6 Ω resistor.
Solution 21: (a) R1 = 6 ohm
R2 = 4 ohm
R = R1 + R2 = 6 + 4 = 10 ohm
V = 20 V
I = V/R = 20/10 = 2 A
(b) R = 6 W
I = 2 A
V = ?
V = IR = 6 × 2 = 12 V
Question 22: In fig., calculate:
(a) the total resistance of the circuit
(b) the value if R, and
(c) the current flowing in R.
Solution 22: For resistor A:
R = 1 ohm
V = 2 V
I = V/R = 2/1 = 2A
For resistor B:
R = 2 ohm
V = 2 V
I = V/R = 2/2 = 1A
Question 23: A wire of length 5 m has a resistance of 2.0Ω calculate:
(a) the resistance of wire of length 1 m
(b) the equivalent resistance if two such wires each of length 2 m are joined in parallel.
(c) the resistance of 1 m length of wire of same material but of half diameter.
Solution 23: (a) V = 4 V
I = 0.4 A
Total Resistance R' = ?
R' = V/I = 0.4/4 = 10 ohm
(b) R1 = 20 ohm
R' = 10 ohm
1/R’ = 1/R + 1/R1
1/10 = 1/R + 1/20
1/R = 1/10 - 1/20 = 1/20
R = 20 Ω
(c) R = 20 ohm
V = 4V
I = V/R = 4/20 = 0.2A
Question 24: A particular resistance wire has a resistance of 3.0 ohm per meter. Find:
(a) The total resistance of three lengths of this wire each 1.5 m long, joined in parallel.
(b) The potential difference of the battery which gives a current od 2.0 A in each of the 1.5 m length when connected in parallel to the battery (assume that the resistance of battery is negligible)
(c) the resistance of 5 m length of a wire of the same material, but with twice the area of cross section.
Solution 24: (a) Resistance of 1m of wire = 3 ohm
Resistance of 1.5 m of wire = 3 x 1.5 = 4.5 W
1/R = 1/4.5 + 1/4.5 + 1/4.5 = 3/4.5
R = 1.5 oh m
V = 4V
I = V/R = 4/20 = 0.2A
Question 24: A particular resistance wire has a resistance of 3.0 ohm per meter. Find:
(a) The total resistance of three lengths of this wire each 1.5 m long, joined in parallel.
(b) The potential difference of the battery which gives a current od 2.0 A in each of the 1.5 m length when connected in parallel to the battery (assume that the resistance of battery is negligible)
(c) the resistance of 5 m length of a wire of the same material, but with twice the area of cross section.
Solution 24: (a) Resistance of 1m of wire = 3 ohm
Resistance of 1.5 m of wire = 3 x 1.5 = 4.5 W
1/R = 1/4.5 + 1/4.5 + 1/4.5 = 3/4.5
R = 1.5 oh m
(b) I = 2 A
V = IR = 2 x 4.5 = 9 V
V = IR = 2 x 4.5 = 9 V
(c) R = 3 ohm for 1 m
For 5 m: R = 3 × 5 = 15 ohm
But Area A is double i.e. 2A and Resistance is inversely proportional to area so Resistance will be half.
R = 15/2 = 7.5 ohm
Question 25: A cell supplies a current of 1.2 A through two 2 Ω resistors connected in parallel. When the resistors are connected in series, it supplies a current of 0.4 A. Calculate: (i) the internal resistance and (ii) e.m.f. of the cell.
Solution 25: In parallel R = 1/2 + 1/2 = 1 ohm
I = 1.2 A
ε = I(R + r) = 1.2(1 + r) = 1.2 + 1.2 r
In series R = 2+2 = 4 ohm
I = 0.4 A
ε = I(R + r) = 0.4(4 + r) = 1.6 + 0.4 r
It means :
1.2 + 1.2 r = 1.6 + 0.4 r
0.8 r = 0.4
R = 0.4 / 0.8 = 1/2 = 0.5 ohm
(i) Internal resistance r = 0.5 ohm
(ii) ε = I(R+r) = 1.2(1+0.5) = 1.8 V
Question 26: A battery of e.m.f 15 V and internal resistance 3 Ω is connected to two resistors 3 Ω and 6 Ω connected in parallel. Find: (a) the current through the battery. (b) p.d. between the terminals of the battery, (c) the current in 3 Ω resistors, (d) the current in 6 Ω resistor.
Solution 26: (a) In parallel 1/R = 1/3 + 1/6 = 1/2
So R = 2 ohm
r = 3 W
ε = 15 V
ε = I(R + r)
15 = I(2 + 3)
I = 15/5 = 3 A
For 5 m: R = 3 × 5 = 15 ohm
But Area A is double i.e. 2A and Resistance is inversely proportional to area so Resistance will be half.
R = 15/2 = 7.5 ohm
Question 25: A cell supplies a current of 1.2 A through two 2 Ω resistors connected in parallel. When the resistors are connected in series, it supplies a current of 0.4 A. Calculate: (i) the internal resistance and (ii) e.m.f. of the cell.
Solution 25: In parallel R = 1/2 + 1/2 = 1 ohm
I = 1.2 A
ε = I(R + r) = 1.2(1 + r) = 1.2 + 1.2 r
In series R = 2+2 = 4 ohm
I = 0.4 A
ε = I(R + r) = 0.4(4 + r) = 1.6 + 0.4 r
It means :
1.2 + 1.2 r = 1.6 + 0.4 r
0.8 r = 0.4
R = 0.4 / 0.8 = 1/2 = 0.5 ohm
(i) Internal resistance r = 0.5 ohm
(ii) ε = I(R+r) = 1.2(1+0.5) = 1.8 V
Question 26: A battery of e.m.f 15 V and internal resistance 3 Ω is connected to two resistors 3 Ω and 6 Ω connected in parallel. Find: (a) the current through the battery. (b) p.d. between the terminals of the battery, (c) the current in 3 Ω resistors, (d) the current in 6 Ω resistor.
Solution 26: (a) In parallel 1/R = 1/3 + 1/6 = 1/2
So R = 2 ohm
r = 3 W
ε = 15 V
ε = I(R + r)
15 = I(2 + 3)
I = 15/5 = 3 A
(b) V = ?
R = 2 ohm
V = IR = 3 x 2 = 6 V
R = 2 ohm
V = IR = 3 x 2 = 6 V
(c) V = 6 V
R = 3 ohm
I = V/R = 6/3 = 2 A
R = 3 ohm
I = V/R = 6/3 = 2 A
(d) R = 6 ohm
V = 6 V
I = V/R = 6/6 = 1 A
Question 27: The following circuit diagram (Fig.) shows three resistors 2 Ω, 4 Ω and R Ω connected to a battery of e.m.f 2 V and internal resistance 3 Ω. A main current of 0.25A flows through the circuit.
V = 6 V
I = V/R = 6/6 = 1 A
Question 27: The following circuit diagram (Fig.) shows three resistors 2 Ω, 4 Ω and R Ω connected to a battery of e.m.f 2 V and internal resistance 3 Ω. A main current of 0.25A flows through the circuit.
(b) Calculate the p.d. across the internal resistance of the cell.
(c) What is the p.d. across the R Ω or 2 Ω resistor?
(d) calculate the value of R.
Solution 27: (a) R = 4 Ω
I = 0.25 A
V = IR = 0.25 x 4 = 1 V
(b) Internal Resistance r = 3 ohm
I = 0.25 A
V = IR = 0.25 x 3 = 0.75 V
I = 0.25 A
V = IR = 0.25 x 3 = 0.75 V
(c) Effective resistance of parallel combination of two 2 ohm resistances = 1 ohm
V = I/R = 0.25/1 = 0.25 V
V = I/R = 0.25/1 = 0.25 V
(d) I = 0.25 A
ε = 2V, r = 3 ohm
ε = I(R' + r)
2 = 0.25(R' + 3)
R' = 5 W
2R/(2 + R) + 4 = 5
R = 2 ohm
Question 28: Three resistors of 6.0 Ω, 2.0 Ω and 4.0 Ω are joined to an ammeter A and a cell of e.m.f. 6.0 V as shown in fig. Calculate:
(a) the effective resistance of the circuit and
(b) the reading of ammeter.
Solution 28: (a) R1 = 6 W
R' = R2 + R3 = 2 + 4 = 6 W
R1 and R' in parallel :
1/R = 1/R1 + 1/R’
= 1/6 + 1/6 = 2/6
R = 3 ohm
ε = 2V, r = 3 ohm
ε = I(R' + r)
2 = 0.25(R' + 3)
R' = 5 W
2R/(2 + R) + 4 = 5
R = 2 ohm
Question 28: Three resistors of 6.0 Ω, 2.0 Ω and 4.0 Ω are joined to an ammeter A and a cell of e.m.f. 6.0 V as shown in fig. Calculate:
(a) the effective resistance of the circuit and
(b) the reading of ammeter.
R' = R2 + R3 = 2 + 4 = 6 W
R1 and R' in parallel :
1/R = 1/R1 + 1/R’
= 1/6 + 1/6 = 2/6
R = 3 ohm
(b) R = 3 ohm
V = 6V
I = ?
I = V/R = 6/3 = 2A
Question 29: The diagram below in Fig. shows the arrangement of five different resistances connected to a battery of e.m.f. 1.8V Calculate:
(a) the total resistance of the circuit, and
(b) the reading of ammeter A.
Solution 29 :
(a) In the figure above,
Let resistance between X and Y be Rxy
Then, 1/Rxy = 1/10 + 1/40 = (4 + 1)/40 = 5/40 Ω
Or, Rxy = 8 Ω
Let RAB be the net resistance between points A and B.
Then, 1/RAB = 1/30 + 1/20 + 1/60 = (2 + 3 + 1)/60 = 6/60 Ω
Or, RAB = 10 Ω
∴ Total resistance of the circuit = 8 Ω + 10 Ω = 18 Ω
V = 6V
I = ?
I = V/R = 6/3 = 2A
Question 29: The diagram below in Fig. shows the arrangement of five different resistances connected to a battery of e.m.f. 1.8V Calculate:
(a) the total resistance of the circuit, and
(b) the reading of ammeter A.
Let resistance between X and Y be Rxy
Then, 1/Rxy = 1/10 + 1/40 = (4 + 1)/40 = 5/40 Ω
Or, Rxy = 8 Ω
Let RAB be the net resistance between points A and B.
Then, 1/RAB = 1/30 + 1/20 + 1/60 = (2 + 3 + 1)/60 = 6/60 Ω
Or, RAB = 10 Ω
∴ Total resistance of the circuit = 8 Ω + 10 Ω = 18 Ω
(b) Current I = Voltage/(Total resistance) = 1.8/18A
Or, I = 0.1 A
Thus, 0.1 A shall be the reading of the ammeter
Or, I = 0.1 A
Thus, 0.1 A shall be the reading of the ammeter