250+ TOP MCQs on Current Electricity – Cells in Series and in Parallel | Class12 Physics

Physics Multiple Choice Questions on “Current Electricity – Cells in Series and in Parallel”.

1. What will be the grouping of cells when the current in the circuit is (frac {ne}{(R + nr)})?
a) Parallel grouping
b) Series grouping
c) Mixed grouping
d) When there is no grouping
Answer: b
Clarification: When n identical cells, each of emf ‘e’ and internal resistance ‘r’ are connected to the external resistance ‘R’ in series, its called series grouping. In series grouping e_eq = ne and r_eq = nr Therefore, current in the circuit (I) = (frac {ne}{(R + nr)}).

2. Which of the following is correct when one cell is wrongly connected in series circuit?
a) The total emf reduces by e
b) The total emf increases by e
c) The total emf increases by 2e
d) The total emf decreases by 2e
Answer: c
Clarification: When one cell is wrongly connected in series of n identical cells, each of emf e, it will reduce the total emf by 2e. So, effective emf is calculated as e_EFF = ne – 2e. This happens in case of mixed grouping.

3. Calculate the number of dry cells, each of emf 2V and internal resistance 1V that is joined in series with a resistance of 30 ohms so that a current of 0.8A passes through it.
a) 20
b) 10
c) 30
d) 40
Answer: a
Clarification: Emf = 2V; r = 1 ohm; I = 0.8A; R = 30 ohms
The required equation: I = (frac {(n times e)}{ [(n times r) + R] })
0.8 = (frac {(n times 2)}{ [(n times 1) + 30] })
0.8n + 24 = 2n
1.2n = 24
n = (frac {24}{1.2}) = 20
Therefore, the number of dry cells required are 20.

4. In parallel grouping of cells, we obtain more current.
a) True
b) False
Answer: a
Clarification: Yes, in parallel grouping of cells, we obtain more current. In parallel combination, the voltage remains the same and the resistance offered is minimum. As more cells are added parallel to each other, then the resistance will keep reducing. As a result, more current can be obtained.

5. There are 4 resistors, each having the same resistance of 4 ohms. These are first connected in series with a cell of internal resistance 2 ohms. Then, they are connected in parallel to the same cell. Find the ratio of the respective currents in the two cases.
a) 1:8
b) 1:7
c) 1:6
d) 6:1
Answer: c
Clarification: When the resistors are connected in series:
R_S = 4 + 4 + 4 + 4 = 16 ohms
They are connected to a cell of internal resistance of 2 ohms, so current (I₁) = (frac {e}{R_S}) + r
= (frac {e}{6}) + 2
= (frac {e}{18})
When the resistors are connected in parallel:
(frac {1}{R_P} = frac {1}{4} + frac {1}{4} + frac {1}{4} + frac {1}{4}) = 1 → R_P = 1
The current through the circuit (I₂) = (frac {e}{1+2})
= (frac {e}{3})
Ratio of both the currents = (frac {I_1}{I_2} = frac {frac {e}{18}}{frac {e}{3}} = frac {1}{6}) → 1:6

6. 36 cells, each of emf 4V are connected in series and kept in a box. The combination shows an emf of 88V on the outside. Calculate the number of cells reversed.
a) 2
b) 5
c) 10
d) 7
Answer: d
Clarification: Number of cells (n) = 36; Emf of each cell (e) = 4V; Total emf (E) = 88V;
Let the number of reversed cells be ‘y’
The required equation: E_EFF = n × e – 2y × e
88 = 36 × 4 – 2y × 4
88 = 144 – 8y
8y = 56
y = 7
Therefore, there are 7 reversed cells.

7. ‘n’ cells have emf ‘e’ and internal resistance ‘r’ and connected to an external resistance ‘R’. They pass the same current whether the cells are connected in series or in parallel to each other. Then which of the following conditions are true?
a) R = r
b) r = nR
c) R = nr
d) R = n²r
Answer: a
Clarification: Current passed through the external resistance when the cells are connected in series:
I₁ = (frac {ne}{(R + nr)})
Current passed through the external resistance when the cells are connected in parallel:
I₂ = (frac {ne}{(nR + r)})
Given: I₁ = I₂
(frac {ne}{(R + nr)} = frac {ne}{(nR + r)})
R + nr = nR + r
R – nR = r – nr
R (1 – n) = r (1 – n)
R = r

8. When cells are connected incorrectly in series, total internal resistance is also affected.
a) True
b) False
Answer: b
Clarification: When ‘n’ cells, each of internal resistance ‘r’, are incorrectly connected in series, the total internal resistance of cells still remains nr, i.e. there is no effect on the total internal resistance of the cells.

9. A battery of emf 10V has an internal resistance of 1 ohms and is charged by a 150V dc supply using a series resistance of 19 ohms. What is the terminal voltage of the battery?
a) 15V
b) 20V
c) 17V
d) 25V
Answer: c
Clarification: Emf (e) = 10; Internal resistance (r) = 1 ohm; DC supply given = 150V; Resistance(R) = 19 ohms E_EFF = 150 – 10 = 140V; R_TOT = R + r = 19 + 1 = 20 ohms.
I = (frac {E_{EFF}}{(R + r)} = frac {140}{20}) = 7A
Terminal voltage = emf of battery + voltage drop across battery
= 10 + Ir
= 10 + (7 x 1)
= 10 + 7 = 17V
Therefore, the terminal voltage of battery is 17V.

10. A cell has an emf ‘e’ and internal resistance ‘r’ and is connected across a variable external resistance R. Identify the correct plot from the following of potential difference across resistance R when R is increased.
a)

b)

c)

d)

Answer: b
Clarification: Current passing in the circuit (I) = (frac {e}{(R + r)})
Potential difference (V) = IR = [ (frac {e}{(R + r)}) ] × R
V = (frac {e}{(1 + frac {r}{R})})
Therefore, when R = 0 → V = 0
R = infinity → V = e
Thus, an upward curve will only be obtained.