Two batteries of e.m.f 4 V and 8 V with internal resistance 1 Ω and 2 Ω respectively are connected in a circuit with a resistance of 9 Ω as shown in the figure. The current and potential difference between the points ' P ' and ' Q ' is R = 9 Ω (a) 1/3 A and 4 V (b) 1/3 A and 3 V (c) 1/2 A and 5 V (d) 1/6 A and 3 V

Answer: (b). The two cells face each other (+ to +), so they oppose: net emf = 8 - 4 = 4 V. Total resistance = 1 + 2 + 9 = 12 Ω, I = 4/12 = 1/3 A. P and Q are the ends of R: VPQ = IR = 1/3 × 9 = 3 V So the answer is (b) 1/3 A and 3 V.

Two batteries of e.m.f 4 V and 8 V with internal resistance 1 Ω and 2… | MHT CET 2025 (25 Apr, Shift 2)

Q28MHT CET 2025 · 25 APR · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments

Two batteries of e.m.f 4 V and 8 V with internal resistance $1\ \Omega$ and $2\ \Omega$ respectively are connected in a circuit with a resistance of $9\ \Omega$ as shown in the figure. The current and potential difference between the points ' P ' and ' Q ' is $R = 9\ \Omega$

Show answer and solution

Answer: (b)

The two cells face each other (+ to +), so they oppose: net emf $= 8 - 4 = 4$ V.
Total resistance $= 1 + 2 + 9 = 12\ \Omega$ , $I = \frac{4}{12} = \frac{1}{3}$ A.
P and Q are the ends of $R$ : $V_{PQ} = IR = \frac{1}{3} \times 9 = 3$ V
So the answer is (b) $\frac{1}{3}$ A and 3 V.

Report a mistake

Kirchhoff's Laws & Measuring Instruments question from MHT CET 2025 (25 Apr, Shift 2)

More questions on Kirchhoff's Laws & Measuring Instruments