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Cells are devices that convert chemical energy into electrical energy. This electrical energy is then used to power up electrical devices. Cells are connected in two ways, i.e. in series connection and parallel connection. Cells in series and parallel connection topic is part of Class 12 Chapter 3 Current Electricity and holds a good amount of weightage in the Class 12 Physics Syllabus.
- Cells in Series: When the cells are connected from end to end in a single line connection, it is called a series connection.
The overall emf (E) of the circuit where cells are connected in series is,
E = E1 + E2 + E3 + …En
Similarly, the overall internal resistance of the entire circuit is,
r = r1 + r2 + r3 +....rn
- Cells in Parallel: When the cells are connected parallelly to one another, it is called a parallel connection.
The overall internal resistance of the entire circuit is,
\(r = ( \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + ............. \frac{1}{r_n})\)From the above equations, we can calculate the overall current flowing through the circuit by the following formula.
\(\frac{E_{eq}}{r_{eq}} = \frac{E_1}{r_1} + \frac{E_2}{r_2}\)Check Out:
MCQs on Combination of Cells in Series and Parallel Connection
Ques. What will be the grouping of cells when the current in the circuit is ne/(R+nr)?
a) Parallel grouping
b) Series grouping
c) Mixed grouping
d) When there is no grouping
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Ans. The correct option is (b)
Explanation: When n identical cells, each of emf ‘e’ and internal resistance ‘r’ is connected to the external resistance ‘R’ in series, it is called a series connection. In series grouping
eeq = ne and req = nr
Therefore,
current in the circuit (I) = ne/(R+nr).
Ques. Which of the following is correct when one cell is wrongly connected in a 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
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Ans. The correct option is (c)
Explanation: When one cell is wrongly connected in a series of n identical cells, each of emf e, it will reduce the total emf by 2e.
So, effective emf is calculated as
eEFF = ne – 2e.
Ques. 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
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Ans. The correct option is (a)
Explanation: Emf = 2V; r = 1 ohm; I = 0.8A; R = 30 ohms
The required equation: I = (n×e)/[(n×r)+R]
0.8 = [(n×2)/[(n×1)+30]
0.8n + 24 = 2n
1.2n = 24
n = 24/1.2 = 20
Therefore, the number of dry cells required is 20.
Ques. In a parallel grouping of cells, we obtain more current.
a) True
b) False
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Ans. The correct option is (a)
Explanation: True, in a parallel connection 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, the resistance will keep reducing. As a result, more current can be obtained.
Ques. 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
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Answer: c
Explanation: When the resistors are connected in series:
RS = 4 + 4 + 4 + 4 = 16 ohms
They are connected to a cell of internal resistance (r) of 2 ohms, so current (I1) = e/(RS + r)
= e/(16 + 2)
= e/18
When the resistors are connected in parallel:
1/RP = 1/4 + 1/4 + 1/4 + 1/4 = 1
⇒ RP = 1
The current through the circuit (I2) = e/(1+2) = e/3
The ratio of both the currents
\(\frac {I_1}{I_2}= \frac {e/18}{e/3}=\frac {1}{6}\)
Ques. 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
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Ans The correct option is (d)
Explanation: Given
- 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: EEFF = (n × e) – (2y × e)
88 = (36 × 4) – (2y × 4)
88 = 144 – 8y
8y = 56
y = 7
Therefore, there are 7 reversed cells.
Ques. ‘n’ cells have emf ‘e’ and internal resistance ‘r’ and are 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
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Ans. The correct option is (a)
Explanation: Current passed through the external resistance when the cells are connected in series: I1 = ne/(R+nr)
Current passed through the external resistance when the cells are connected in parallel:
I2 = ne/(nR+r)
I1 = I2 (given)
ne/(R+nr) = ne/(nR+r)
R + nr = nR + r
R – nR = r – nr
R (1 – n) = r (1 – n)
R = r
Ques. When cells are connected incorrectly in series, total internal resistance is also affected.
a) True
b) False
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Ans. The correct option is (b)
Explanation: 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.
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Ques. A battery of emf 10V has an internal resistance of 1 ohm 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
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Ans. The correct option is (c)
Explanation: Given
- Emf (e) = 10;
- Internal resistance (r) = 1 ohm;
- DC supply given = 150V;
- Resistance(R) = 19 ohms
- EEFF = 150 – 10 = 140V;
- RTOT = R + r = 19 + 1 = 20 ohms.
I = EEFF/(R+r)=140/20 = 7A
Terminal voltage = emf of battery + voltage drop across the battery
= 10 + Ir
= 10 + (7 x 1)
= 10 + 7 = 17V
Therefore, the terminal voltage of the battery is 17V.
Ques. What is the internal resistance of a cell of emf 1.5 V, if it can deliver a maximum current of 3 A?
a) 0.5 Ω
b) 4.5 Ω
c) 2 Ω
d) 1 Ω
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Ans. The correct option is (a)
Explanation: For the maximum amount, load resistance = 0
⇒ E = Ir
r = E/I = 1.5/ 3
= 0.5 Ω
Ques. The terminal voltage of a given cell depends on
a) External resistance, Internal Resistance
b) External resistance
c) Internal Resistance
d) None of these
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Ans. The correct option is (a)
Explanation: Some of the energy of the cell is lost due to the cell’s internal resistance. Thus the total potential difference is reduced by the factor of the internal and external resistance.
Ques. Five cells are connected in a parallel combination with an emf of 2V each. The output voltage is?
a) 10V
b) 1.5V
c) 2V
d) 2.5V
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Ans. The correct option is (c)
Explanation: From the question, all the cells are connected parallel to each other.
Thus,
Equivalent voltage = E
Therefore,
E = 2V
Ques. Choose the correct statement from the following.
a) emf is maximum and internal resistance is minimum in a parallel combination.
b) emf is minimum and internal resistance is maximum in a parallel combination.
c) emf is maximum and internal resistance is minimum in a series combination.
d) Both emf and internal resistance are maximum in a series combination.
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Ans. The correct option is (d)
Explanation: In series combination, both emf and internal resistance will be maximum. On the other hand, in parallel combination, the internal resistance is minimum in a parallel circuit.
Ques. Four cells are connected in a parallel combination with an emf of 5V each. The output voltage is?
a) 10V
b) 1.5V
c) 2V
d) 5V
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Ans. The correct option is (d)
Explanation: From the question, all the cells are connected parallel to each other.
Thus,
Equivalent voltage = E
Therefore,
E = 5V
Ques. Two cells 1.25V and 0.75V are connected in a parallel combination. The effective voltage is?
a) 10V
b) 0.5V
c) 2V
d) 5V
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Answer: The correct option is (b)
Explanation: Assume Internal resistance is zero.
By applying Kirchhoff’s law, we have,
Veff = E1 - E2
E1 = 1.25V
E2 = 0.75V
Veff = 1.25 - 0.75
= 0.5V
Ques. A battery of emf 20 V has an internal resistance of 2 ohms and is charged by a 100 V dc supply using a series resistance of 15 ohms. What is the terminal voltage of the battery?
a) 21 V
b) 22.6 V
c) 29.4 V
d) 25.5 V
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Ans. The correct option is (c)
Explanation: Given
- Emf (e) = 20 V
- Internal resistance (r) = 2 ohm;
- DC supply given = 100 V;
- Resistance(R) = 15 ohms
- EEFF = 100 – 20 = 80 V;
- RTOT = R + r = 15 + 2 = 17 ohms.
I = EEFF/(R+r) = 80/17 = 4.7 A
Terminal voltage = emf of battery + voltage drop across the battery
= 20 + Ir
= 20 + (4.7 x 2)
= 20 + 9.4 = 29.4 V
Therefore, the terminal voltage of the battery is 29.4 V.
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