Exercise 8(b)
Question: 1 Explain the meaning of the terms e.m.f., terminal voltage and internal resistance of cell.
Solution: E.m.f : When no current is drawn from the cell i.e. the cell is in open circuit, the potential difference between the
terminals of the cell is called its electro-motive force or e.m.f.
Terminal voltage: When no current is drawn from the cell i.e. the cell is in closed circuit, the potential difference between
the electrodes of the cell is known as terminal voltage.
Internal resistance: The resistance offered by the electrolyte inside the cell, to the flow of current, is known as the
internal resistance of the cell.
Question: 2 State two differences between the e.m.f. and terminal voltage of a cell. Solution:
e.m.f of cell Terminal voltage of cell
1. It is the characteristic of the cell, i.e., it does not depend on 1. It depends on the amount of current drawn from the cell. More
the amount of current drawn from the cell. the current drawn from the cell, less is the terminal voltage.
2. It is equal to the terminal voltage when the cell is not in use, 2. It is equal to the e.m.f. of cell when cell is not in use, while less
while greater than the terminal voltage when cell is in use. than the e.m.f. when cell is in use.
Question: 3 Name two factors on which the internal resistance of a cell depends and state how does it depend on the
factors stated by you.
Solution: The factors on which the internal resistance of a cell depends are:
(i) The surface area of the electrodes – larger the surface area of 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. ε 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: (a) The total resistance of circuit = R + r
(b) The current drawn from the cell We know that, ε = V + v = IR + Ir = I (R + r)
I = ε / (R + r)
(c) p.d. across the cell: [ε / (R + r)] × R (d) voltage drop inside the cell: [ε / (R + r)] × r
Question: 5 A cell is used to send current to an external circuit. (a) How does the voltage across its terminal compare
with its emf? (b) Under what condition is the emf of the cell equal to its terminal voltage?
Solution: (a) When current is drawn from a cell, its terminal voltage V is less than its e.m.f. ∴ Terminal voltage < e.m.f.
(b) When no current is drawn, then the e.m.f. is equal to the terminal voltage.
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: The p.d. across the terminals of a cell is more in an open circuit and reduced in a closed circuit because in
closed circuit some of the energy is used up in overcoming internal resistance of a cell.
Q 7 Write the expressions for the equivalent resistance R of three resistors R1, R2 and R3 joined in (a) parallel, (b) series.
Solution: (a) Total resistance in parallel is given by: 1 / R = 1 / R1 + 1 / R2 + 1 / R3
(b) Total resistance in series is given by R = R1 + R2 + R3
Question: 8 How would you connect two resistors in series? Draw a diagram. Calculate the total equivalent resistance.
Solution: If current I is drawn from the battery, the current will also be I through each
resistor. Applying Ohm’s law to the two resistors separately, we get,
V1 = IR1 V2 = IR2
V = V 1 + V2
IR = IR1 + IR2
R = R1 + R2 Thus total resistance in series is 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 the combination.
Solution: Applying Ohm’s law separately to the two resistors, we have
I1 = V / R1 I2 = V / R2
I = I1 + I2 V / R = V / R1 + V / R 2
1 / R = 1 / R 1 + 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: (a) The two resistors are in series. (b) The two resistors are in parallel.
(c) The two resistors are in parallel. (d) The two resistors are in 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: Change in V is less for the straight line A than for the straight line B (which
means the straight line A is less steeper than B). Hence, the straight line A
represents small resistance and the straight line B represents more resistance. The
resistance decreases in parallel combination while the resistance increases in series
combination. Thus the straight line A represents the parallel combination.
NUMERICALS
Question: 1 The diagram in figure shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm connected 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 ammetere and
voltmeter when (i) the key K is open, and (ii) the key K is closed
Solution: (i) Because of no current, ammeter reading = 0
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 × 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 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: Given, 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 emf 1.8 V and internal resistance 2 ohm is connected in series with
an ammeter of resistance 0.7 ohm and resistance of 4.5 ohm as shown in figure.
(a) What would be the reading of the ammeter?
(b) What is the potential difference across the terminals of the cell?
Solution: (a) ε = 1.8 V Total resistance = 2 + 4.5 + 0.7 = 7.2 W I =?
I = ε / R (total resistance) = 1.8 / 7.2 = 0.25 A
(b) Current I = 0.25 A [calculated in (a) part] Now, excluding internal resistance total resistance = 4.5 + 0.7 = 5.2 ohm
V = IR = 0.25 × 5.2 = 1.3 V
Question: 4 The music system draws a current of 400 mA when connected to a 12 V battery.
(a) What is the resistance of the music system?
(b) The music system if left playing for several hours and finally the battery voltage drops and the music system stops
playing when the current drops to 320 mA. At what battery voltage does the music system stop playing?
Solution: V = IR R = V / I = 12 / 400 mA = 12*0.4 = 30 ohm.
I’ = 320 mA , R = 30 ohm
V’ = I’R = 320 mA = 0.32 * 30 = 9.6 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 ohm. But its sends current 0.5 A when it is connected to an external resistance of 3.9 ohm. Calculate the values
of e and r.
Solution: In first case, I = 1 A, R = 1.9 ohm ε = I (R + r) = 1 (1.9 + r) = 1.9 + r [1]
In second case, I = 0.5 A, R = 3.9 ohm ε = I (R + r) = 0.5 (3.9 + r) = 1.95 + 0.5r [2]
From equation [1] and [2] 1.9 + r = 1.95 + 0.5r
r = 0.05 / 0.5 = 0.1 ohm
Now, substituting the value of r ε = 1.9 + r = 1.9 + 0.1 = 2 V
Q:6 Two resistors having resistance 4 ohm and 6 ohm are connected in parallel. Find their equivalent resistance.
Solution: Let the equivalent resistance of the 4 ohm and 6 ohm resistors connected in parallel be R’
Then, 1 / R’ = 1 / 4 + 1 / 6 = (3 + 2) / 12 = 5 / 12 ohm or R’ = 12 / 5 = 2.4 ohm
Question: 7 Four resistors each of resistance 2 ohm are connected in parallel. What is the effective resistance?
Solution: R1 = 2 ohm, R2 = 2 ohm, R3 = 2 ohm, R4 = 2 ohm
1 / R = 1 / R 1 + 1 / R2 + 1 / R3 + 1 / R4
1/R=1/2+1/2+1/2+1/2 1/R=2 so 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: To get a total resistance less than 1 Ω, the three resistors should
be connected in parallel
Let the total resistance be R’ Then, 1 / R’ = 1 / 2 + 1 / 3 + 1 / 5
1 / R’ = (15 + 10 + 6) / 30
1 / R’ = 31 / 30 Ω or
R’ = 30 / 31 = 0.97 Ω
Question: 9 Three resistors each of 2 ohm are connected together so that their total resistance is 3 ohm. Draw a
diagram to show this arrangement and check it by calculation.
Solution: 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’ = 1 / 2 + 1 / 2 1 / R’ = 1 so, R’ = 1 ohm
R = R’ + R3 = 1 + 2 = 3 ohm
Question: 10 Calculate the equivalent resistance between the points A and B in figure if each resistance is 2.0 Ω
Solution: For a parallel resistances
Reff = (R1R2) / (R1 + R2)
Reff = (2 × 2) / (2 + 2) = 4 / 4 = 1 Ω
Hence, total resistance = 2 + 2 + 1 = 5 Ω
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: 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, the above 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
1/r=4/6 r = 6 / 4 = 1.5 ohm
Question: 12 In the circuit shown below in figure, calculate the value of x
if the equivalent resistance between the points A and B is 4 ohm
Solution: r1 = 4 ohm, r2 = 8 ohm, r3 = x ohm, r4 = 5 ohm
r = 4 ohm r’ = r1 + r2 = 4 + 8 = 12 ohm
r’’ = r3 + r4 = (x + 5) ohm
1 / r = 1 / r’ + 1 / r’’
1 / 4 = 1 / 12 + 1 / (5 + x)
1 / 4 – 1 / 12 = 1 / (5 + x)
(3 – 1) / 12 = 1 / (5 + x)
2 / 12 = 1 / (5 + x)
1 / 6 = 1 / (5 + x) x = 1 ohm
Q 13 Calculate the effective resistance between the points A and B in the circuit shown in figure
Solution:
In above figure,
Resistance between XAY = (1 + 1 + 1) = 3 ohm
Resistance between XY = 2 ohm
Resistance between XBY = 6 ohm
Let the net resistance between the points X and Y be R’
Then, 1 / R’ = 1 / 2 + 1 / 3 + 1 / 6
1 / R’ = (3 + 2 + 1) / 6
1 / R’ = 6 / 6
1 / R’ = 1 ohm or R’ = 1 ohm
Therefore, we can say that three 1 ohm resistors are connected in series
between points A and B.
Let the net resistance between points A and B be RAB
Then, RAB = (1 + 1 + 1) ohm = 3 ohm
Question: 14 A uniform wire with a resistance of 27 ohm is divided into three equal pieces and then they are joined in
parallel. Find the equivalent resistance of the parallel combination.
Solution: Since the wire is divided into three pieces,
The new resistance = 27 / 3 = 9
Now, three resistance are joined in parallel 1 / r = 1 / r1 + 1 / r2 + 1 / r3
1/r=1/9+1/9+1/9
1/r=3/9
1/r=1/3 r = 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 of the circuit. Draw a diagram of the arrangement.
Solution: 1 / r = 1 / 6 + 1 / 3 1/r=1/2 r = 2 ohm
R=2+1 = 3 ohm
Question: 16 Calculate the effective resistance between the points A and B in the
network shown below in figure.
Solution: For parallel resistance 1 / R = 1 / 12 + 1 / 6 + 1 / 4
1 / R = (1 + 2 + 3) / 12 = 6 / 12
R = 12 / 6 = 2 ohm
Now, all the resistances are in series R = 2 + 2 + 5 = 9 ohm
Q17 Calculate the equivalent resistance between the points A and B in figure
Solution: Given, R1 = 3 + 2 = 5 ohm R2 = 30 W R3 = 6 + 4 = 10 ohm
The resistors R1, R2 and R3 are connected in parallel
1 / R = 1 / R 1 + 1 / R 2 + 1 / R3
1 / R = 1 / 5 + 1 / 30 + 1 / 10
1 / R = (6 + 1 + 3) / 30
1 / R = 10 / 30
1/R=1/3 R = 3 ohm
Question: 18 In the network shown in adjacent figure, calculate the
equivalent resistance between the points (a) A and B (b) A and C
Solution: (a) R1 = 2 + 2 + 2 R1 = 6 ohm R2 = 2 ohm
R1 and R2 are connected in parallel 1 / R = 1 / R 1 + 1 / R2
1/R=1/6+1/2
1 / R = (1 + 3) / 6
1/R=4/6
R = 6 / 4 = 1.5 ohm
(b) R1 = 2 + 2 R1 = 4 ohm R2 = 2 + 2 = 4 ohm
The resistors R1 and R2 are connected in parallel 1 / R = 1 / R 1 + 1 / R2
1/R=1/4+1/4
1/R=2/4
1/R=1/2 R = 2 ohm
Question: 19 Five resistors, each of 3 ohm, are connected as shown in figure. Calculate the resistance (a) between the
points P and Q, and (b) between the points X and Y.
Solution: (a) R1 = 3 + 3 R1 = 6 ohm R2 = 3 ohm
R1 and R2 are connected in parallel 1 / R = 1 / R1 + 1 / R2
1/R=1/6+1/3
1 / R = (1 + 2) / 6
1/R=3/6
1/R=1/2 R = 2 ohm
(b) We know that R = 2 ohm from the above calculation R3 = 3 ohm R4 = 3 ohm
R’ = R + R3 + R4 =2+3+3 = 8 ohm
Question: 20 Two resistors of 2 ohm and 3 ohm 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: (a) Here, R1 and R2 are connected in series
R1 = 2 ohm R2 = 3 ohm R = R 1 + R2
R = 2 + 3 = 5 ohm
V=6V Now, I = V / R I=6/5 I = 1.2 ohm
(b) Here, R1 and R2 are connected in parallel 1 / R = 1 / R1 + 1 / R2
1/R=1/2+1/3
1 / R = (3 + 2) / 6
1/R=5/6
R = 6 / 5 = 1.2 ohm
V = 6 V We know that, I = V / R = 6 / 1.2 = 5 A Therefore in series: 1.2 A and in parallel: 5 A
Q21 A resistor of 6 ohm is connected in series with another resistor of 4 ohm. A potential difference of 20 V is applied
across the combination. (a) Calculate the current in the circuit and (b) potential difference across the 6 ohm resistor.
Solution: (a) To calculate current in the circuit R1 = 6 ohm R2 = 4 ohm R = R1 + R2 = 6 + 4 = 10 ohm
V = 20 V I = V / R = 20 / 10 = 2 A
(b) To calculate the potential difference across the 6 ohm resistor R = 6 ohm I=2A V =? V = IR
V = 6 × 2 = 12 V
Q22 Two resistors of resistance 4 Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.
(a) Draw a labeled diagram of the arrangement (b) Calculate current in each resistor.
Solution: (a) Circuit diagram
(b) Equivalent resistance of the circuit 1/R=1/4+1/6
1 / R = (3 + 2) / 12
1 / R = 5 / 12
R = 12 / 5 = 2.4 ohm
Thus, the e.m.f. of the cell is V = IR = 0.5 × 2.4 = 1.2 V
∴ Current through each resistor is I4 = V / R4 = 1.2 / 4 = 0.3 A
I6 = V / R6 = 1.2 / 6 = 0.2 A Hence, 0.3 A in 4 ohm and 0.2 A in 6 ohm
Question: 23 Calculate the current flowing through each of the resistors A and B in
the circuit shown in figure?
Solution: For resistor A R = 1 ohm V=2V
I=V/R =2/1 =2A
For resistor B, R = 2 ohm V=2V I=V/R =2/2 =1A
Hence, current flowing in resistor A is 2 A and current flowing in resistor B is 1 A
Question: 24 In figure, calculate: (a) the total resistance of the circuit.
(b) the value of R, and (c) the current flowing in R
Solution: (a) To calculate the total resistance of the circuit
V=4V I = 0.4 A Total resistance R’ =?
R’ = V / I = 0.4 / 4 = 10 ohm
(b) To calculate the value of R 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 / R = (2 – 1) / 20
1 / R = 1 / 20 R = 20 ohm
(c) To calculate the current flowing in R R = 20 ohm V=4V
I = V / R = 4 / 20 = 0.2 A
Question: 25 A particular resistance wire has a resistance of 3 ohm per meter. Find: (a) The total resistance of three
lengths of this wire each 1.5 m long, in parallel.
(b) The potential difference of the battery which gives a current of 2 A in each of the 1.5 m length when connected in the
parallel to the battery (assume that resistance of the 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: (a) Resistance of wire per meter = 3 ohm
So, resistance of three lengths of this wire each 1.5 m long = 3 × 1.5 = 4.5 W
1 / R = 1 / 4.5 + 1 / 4.5 + 1 / 4.5 = 3 / 4.5 R = 1.5 ohm
(b) I = 2 A V = IR = 2 × 4.5 = 9 V
(c) R = 3 ohm for 1 meter wire For 5 m R = 3 × 5 = 15 ohm
Here the area is twice i.e 2 A and Resistance is inversely proportional to area. Thus resistance becomes half R = 15 / 2
R = 7.5 ohm
Question: 26 A cell supplies a current of 1.2 A through two 2 ohm resistors connected in parallel. When 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: In parallel R = 1 / 2 + 1 / 2 = 1 ohm I = 1.2 A ε = I (R + r) = 1.2 (1 + r) = 1.2 + 1.2r
In series R = 2 + 2 = 4 ohm
I = 0.4 A ε = I (R + r) = 0.4 (4 + r) = 1.6 + 0.4r
This means: 1.2 + 1.2r = 1.6 + 0.4r
1.2r – 0.4r = 1.6 – 1.2
0.8r = 0.4
r = 0.4 / 0.8 = 0.5 ohm
(i) Internal resistance r = 0.5 ohm (ii) ε = I(R + r) = 1.2 (1 + 0.5) = 1.8 V
Question: 27 A battery of emf 15 V and internal resistance 3 ohm is connected to two resistors 3 ohm and 6 ohm
connected in parallel. Find (a) the current through the battery (b) p.d. between the terminals of the battery (c) the
current in 3 ohm resistor (d) the current in 6 ohm resistor.
Solution: (a) In parallel 1 / R = 1 / 3 + 1 / 6 = (2 + 1) / 6 = 3 / 6 = 1 / 2
R = 2 ohm
r = 3 ohm ε = 15 V ε = I (R + r)
15 = I (2 + 3)
I = 15 / 5 = 3 A
(b) V =? R = 2 ohm V = IR =3×2 =6V
(c) V = 6 V R = 3 ohm I=V/R =6/3 =2A
(d) R = 6 ohm V=6V I=V/R I=6/6 =1A
Question: 28 The circuit diagram in figure shows three resistors 2 ohm, 4 ohm and R ohm connected to a battery of e.m.f.
2 V and internal resistance 3 ohm. If main current of 0.25 A flows through the circuit, find:
(a) the p.d. across the 4 ohm resistor
(b) the p.d. across the internal resistance of the cell,
(c) the p.d. across the R ohm or 2 ohm resistor, and
(d) the value of R.
Solution: (a) To calculate the p.d. across the 4 ohm resistor
R = 4 ohm I = 0.25 A V = IR V = 0.25 × 4 = 1 V
(b) To calculate the p.d. across the internal resistance of the cell Internal resistance r = 3 ohm
I = 0.25 A V = IR = 0.25 × 3 = 0.75 V
(c) To calculate the p.d. across the 2 ohm resistor
Effective resistance of parallel combination of 2 ohm resistances = 1 ohm V = I / R = 0.25 / 1 = 0.25 V
(d) To calculate the value of R 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
Q9 Three resistors of 6.0 ohm, 2.0 ohm and 4.0 ohm are joined to an ammeter A and a cell of
emf 6.0 V as shown in figure. Calculate: (a) the effective resistance of the circuit.
(b) the reading of ammeter
Solution: (a) R1 = 6 ohm R’ = R2 + R3 = 2 + 4 = 6 ohm
R1 and R’ are connected in parallel
1 / R = 1 / R1 + 1 / R’ = 1 / 6 + 1 / 6 = 2 / 6 = 1 / 3 R = 3 ohm
(b) R = 3 ohm V=6V I =? I=V/R=6/3=2A
Q30 The diagram below in Fig., shows the arrangement of five different
resistances connected to a battery of e.m.f. 1.8 V. Calculate:
a. The total resistance of the circuit b. The reading of ammeter A.
Solution: (a) In the above figure,
Let RXY be the resistance between X and Y
Then, 1 / RXY = 1 / 10 + 1 / 40 = (4 + 1) / 40 = 5 / 40 ohm
Or RXY = 8 ohm
Let the net resistance between points A and B be RAB
Then, 1 / RAB = 10 ohm
∴ The total resistance of the circuit = 8 ohm + 10 ohm = 18 ohm
(b) Current I = Voltage / Total resistance
I = 1.8 / 18 A = 0.1 A Hence, the reading of ammeter is 0.1 A
Q31 A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an
ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as
shown in fig.
Find: (a) The reading of the ammeter,
(b) The potential difference across the terminals of the cells, and
(c) The potential difference across the 4.5 ohm resistor.
Solution: The total resistance of the circuit is
Req = Rcell + Rammeter + R1 || R2
∴ Req = 1.2 + 0.8 + (R1R2) / R1 + R2
∴Req = 2 + (4.5 × 9) / 4.5 + 9 = 2 + 40.5 / 13.5 = 5 ohm
(a) The current through the ammeter is I = Ecell / Req = 2 / 5 = 0.4 A
(b) The potential difference across the ends of the cells is
Vcell = Ecell – IRcell = 2 – 0.4 × 1.2 = 2 – 0.48 = 1.52 V
(c) The potential difference across the 4.5 ohm resistor is V4.5 = Vcell – Vammeter = 1.52 – 0.4 × 0.8 = 1.52 – 0.32= 1.2 V
