Case: Read the passage and answer the question.
A cell of unknown emf E and internal resistance r, two unknown resistances \( R_1 \) and \( R_2 \) \( (R_2\gt R_1) \) and a perfect ammeter are given. The current in the circuit is measured in five different situations:
(i) without any external resistance in the circuit,
(ii) with resistance \( R_1 \) only,
(iii) with resistance \( R_2 \) only,
(iv) with both \( R_1 \) and \( R_2 \) used in series combination, and
(v) with both \( R_1 \) and \( R_2 \) used in parallel combination.
The currents obtained in the five cases are 0.42 A, 0.6 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in that order.
Match List-I with List-II.
| List-I | List-II |
|---|---|
| (A) With \(R_1\) only | (I) 0.42 A |
| (B) With \(R_2\) only | (II) 0.6 A |
| (C) With \(R_1\) and \(R_2\) in series | (III) 1.05 A |
| (D) With \(R_1\) and \(R_2\) in parallel | (IV) 1.4 A |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
The same cell is used every time, and the current is \( I=E/(R+r) \). So a larger external resistance \( R \) gives a smaller current. The external resistances are in the order: parallel combination < \( R_1 \) < \( R_2 \) < series combination.
(A) With \( R_1 \) only — This matches (III) 1.05 A. \( R_1 \) is larger than the parallel value but smaller than \( R_2 \), so its current is the second largest of the four.
(B) With \( R_2 \) only — This matches (II) 0.6 A. Since \( R_2\gt R_1 \), the current is smaller than with \( R_1 \) alone.
(C) With \( R_1 \) and \( R_2 \) in series — This matches (I) 0.42 A. \( R_1+R_2 \) is the largest external resistance, so the current is the smallest.
(D) With \( R_1 \) and \( R_2 \) in parallel — This matches (IV) 1.4 A. The parallel resistance is smaller than both \( R_1 \) and \( R_2 \), so the current is the largest of the four.
NCERT (2026-27):
Cells, Emf, Internal Resistance – “We also observe that since V is the potential difference across R, we have from Ohm’s law V = I R …”
\( \displaystyle IR=\varepsilon-Ir \quad \text{or,} \quad \)\( \displaystyle I=\frac{\varepsilon}{R+r} \)\( \qquad (3.40) \)