physics : class12 Cell Combination Analysis : Potentiometer : Current Electricity MCQ Questions and Answers
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Cell Combination Analysis
MCQ questions and answers.
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MCQ Questions
1. Two cells of emf 2.0 V and 1.0 V are connected first in series aiding and then in series opposing. If the balance length for the aiding combination is 90 cm, what will be the balance length for the opposing combination?
A.
10 cm
B.
20 cm
C.
30 cm
D.
45 cm
Correct Answer:
30 cm
Explanation:
For the same potentiometer potential gradient, balance length is proportional to emf. Thus, l_a/l_o = (E1 + E2)/(E1 - E2) = 3/1. Hence l_o = 90/3 = 30 cm.
2. Two cells having emfs 2.0 V and 1.0 V and internal resistances 1.0 ohm and 2.0 ohm, respectively, are connected in parallel with their positive terminals together. What emf will be indicated by a potentiometer connected across the combination?
A.
1.20 V
B.
1.50 V
C.
1.67 V
D.
2.00 V
Correct Answer:
1.67 V
Explanation:
For parallel cells of unequal emf, the open-circuit effective emf is (E1/r1 + E2/r2)/(1/r1 + 1/r2). Substitution gives (2/1 + 1/2)/(1 + 1/2) = 5/3 V, or approximately 1.67 V.
3. Cells of emf 1.5 V and 2.0 V, having internal resistances 0.5 ohm and 1.0 ohm, respectively, are connected first in series aiding and then in series opposing. What is the ratio of their potentiometer balance lengths in these two arrangements?
A.
3:1
B.
5:1
C.
7:1
D.
9:1
Correct Answer:
7:1
Explanation:
The potentiometer measures the net emf in each arrangement. For aiding, E_a = 1.5 + 2.0 = 3.5 V; for opposing, E_o = 2.0 - 1.5 = 0.5 V. Therefore, l_a:l_o = 3.5:0.5 = 7:1.
4. A set of n identical cells, each of emf E and internal resistance r, is connected either in series or in parallel to a resistance R. For what value of R will the terminal voltage be the same in both arrangements?
A.
r/n
B.
r
C.
nr
D.
n^2r
Correct Answer:
r
Explanation:
The terminal voltage in series is nER/(R + nr), while in parallel it is ER/(R + r/n). Equating them gives (n - 1)(R - r) = 0. For n greater than 1, R = r.
5. Two cells are connected in series. When connected in series aiding, their potentiometer balance length is 80 cm, and when connected in series opposing, it is 20 cm. What is the ratio of their emfs?
A.
2:1
B.
3:1
C.
5:3
D.
4:1
Correct Answer:
5:3
Explanation:
Let the emfs be E1 greater than E2. Then E1 + E2 is proportional to 80 and E1 - E2 is proportional to 20. Therefore, E1/E2 = (80 + 20)/(80 - 20) = 100/60 = 5/3.
6. Three identical cells, each of emf E and internal resistance r, are arranged with two cells in parallel and this parallel combination in series with the third cell, all in the same direction. What is the ratio of the potentiometer balance length for this arrangement to that for all three cells connected in series?
A.
1:2
B.
2:3
C.
3:4
D.
1:1
Correct Answer:
2:3
Explanation:
The two parallel cells have emf E, and their series combination with the third cell has emf 2E. Three cells in series have emf 3E. Since balance length is proportional to emf, the ratio is 2E:3E = 2:3.
7. Two cells of emf 1.2 V and 2.4 V, with internal resistances 0.4 ohm and 0.8 ohm, respectively, are connected in parallel with like terminals together. What open-circuit emf will a potentiometer measure across the combination?
A.
1.20 V
B.
1.44 V
C.
1.60 V
D.
1.80 V
Correct Answer:
1.60 V
Explanation:
The effective emf is (E1/r1 + E2/r2)/(1/r1 + 1/r2). Thus, it equals (1.2/0.4 + 2.4/0.8)/(1/0.4 + 1/0.8) = 6/3.75 = 1.60 V.
8. Two identical cells are connected in series aiding. Their open-circuit balance length on a potentiometer is 100 cm. When a 5 ohm resistor is connected across the combination, the balance length becomes 80 cm. What is the internal resistance of each cell?
A.
0.25 ohm
B.
0.50 ohm
C.
0.625 ohm
D.
1.25 ohm
Correct Answer:
0.625 ohm
Explanation:
The loaded terminal voltage is 80/100 = 0.8 times the open-circuit emf. For two series cells, V/E_total = R/(R + 2r). Hence 0.8 = 5/(5 + 2r), giving 2r = 1.25 and r = 0.625 ohm.
9. Two cells of emfs 2.0 V and 1.5 V and internal resistances 0.5 ohm and 1.0 ohm, respectively, are connected in parallel with like terminals together. What current circulates between the cells when no external load is connected?
A.
0.20 A
B.
0.33 A
C.
0.50 A
D.
0.67 A
Correct Answer:
0.33 A
Explanation:
With no external load, the circulating current is caused by the emf difference and is I = (E1 - E2)/(r1 + r2). Therefore, I = (2.0 - 1.5)/(0.5 + 1.0) = 0.33 A.
10. A cell of emf E and internal resistance r is connected in series aiding with a combination of two identical cells, each of emf E and internal resistance r connected in parallel. What fraction of the balance length obtained with three identical cells in series will be obtained for this arrangement?
A.
1/3
B.
1/2
C.
2/3
D.
3/4
Correct Answer:
2/3
Explanation:
The parallel pair contributes emf E, and the series cell contributes another E, so the combination has emf 2E. Three cells in series have emf 3E. Since potentiometer balance length is proportional to emf, the fraction is 2E/3E = 2/3.
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