physics : class12 Internal Resistance Measurement : Potentiometer : Current Electricity MCQ Questions and Answers
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Internal Resistance Measurement
MCQ questions and answers.
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MCQ Questions
1. In the potentiometer method for measuring the internal resistance of a cell, the first balancing length is measured when the cell is connected to:
A.
An external resistance only
B.
An open circuit
C.
A short circuit
D.
Another potentiometer
Correct Answer:
An open circuit
Explanation:
With the external circuit open, no current is drawn from the cell, so the balancing length corresponds to its emf.
2. When a known resistance R is connected across a cell, the balancing length of the potentiometer is l. If the balancing length for the open-circuit cell is L, the internal resistance r of the cell is:
A.
r = Rl/L
B.
r = R(L - l)/l
C.
r = Rl/(L - l)
D.
r = R(L + l)/l
Correct Answer:
r = R(L - l)/l
Explanation:
The emf E is proportional to L and the terminal voltage V is proportional to l. Using r = R(E/V - 1), we get r = R(L - l)/l.
3. A cell has an open-circuit balancing length of 100 cm. When a resistance of 4 ohms is connected across it, the balancing length becomes 80 cm. What is the internal resistance of the cell?
A.
0.5 ohm
B.
1 ohm
C.
2 ohms
D.
4 ohms
Correct Answer:
1 ohm
Explanation:
Using r = R(L - l)/l, r = 4(100 - 80)/80 = 1 ohm.
4. Why does the balancing length decrease when a known resistance is connected across the cell?
A.
The terminal voltage becomes less than the emf
B.
The potentiometer wire becomes shorter
C.
The resistance of the known resistor becomes zero
D.
The emf of the cell becomes zero
Correct Answer:
The terminal voltage becomes less than the emf
Explanation:
When current flows through the cell, a voltage drop occurs across its internal resistance, so its terminal voltage is less than its emf and the balancing length decreases.
5. In the internal resistance measurement using a potentiometer, the known resistance R is connected:
A.
In series with the potentiometer wire
B.
Across the terminals of the cell
C.
Only across the galvanometer
D.
Between the jockey and the battery
Correct Answer:
Across the terminals of the cell
Explanation:
Connecting R across the cell draws current from it and allows measurement of the terminal voltage.
6. A cell gives a balancing length of 60 cm when supplying current through a 3-ohm resistance and 75 cm on open circuit. What is its internal resistance?
A.
0.5 ohm
B.
0.75 ohm
C.
1 ohm
D.
1.5 ohms
Correct Answer:
0.75 ohm
Explanation:
Using r = R(L - l)/l, r = 3(75 - 60)/60 = 0.75 ohm.
7. For a cell with internal resistance r and an external resistance R, the terminal voltage V is related to the emf E by:
A.
V = E + Ir
B.
V = E - Ir
C.
V = Ir - E
D.
V = E/R
Correct Answer:
V = E - Ir
Explanation:
The terminal voltage is the emf minus the potential drop Ir across the internal resistance of the cell.
8. If the open-circuit balancing length is twice the balancing length obtained with an external resistance R, the internal resistance of the cell is:
A.
R/2
B.
R
C.
2R
D.
3R
Correct Answer:
R
Explanation:
Since L = 2l, r = R(L - l)/l = R(2l - l)/l = R.
9. If the terminal voltage of a cell is equal to its emf during a potentiometer experiment, the current drawn from the cell is:
A.
Maximum
B.
Zero
C.
Equal to the short-circuit current
D.
Half of the normal current
Correct Answer:
Zero
Explanation:
The terminal voltage equals the emf only when there is no voltage drop across the internal resistance, which means no current is drawn.
10. A cell has an emf of 2 V and a terminal voltage of 1.6 V when connected to a 4-ohm resistance. What is its internal resistance?
A.
0.5 ohm
B.
1 ohm
C.
1.5 ohms
D.
2 ohms
Correct Answer:
1 ohm
Explanation:
The current is I = V/R = 1.6/4 = 0.4 A. Using E = V + Ir, r = (E - V)/I = (2 - 1.6)/0.4 = 1 ohm.