Question

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

• A. $\frac{1}{3}A$ and 3 V
• B. $\frac{1}{6}A$ and 4 V
• C. $\frac{1}{3}A$ and 9 V
• D. $\frac{1}{2}A$ and 12 V

Answer 1

Answer: (A)

$\frac{1}{3}A$ and 3 V

Answer 2

We are given a two-branch circuit between P and Q. One branch consists of a series combination of a 1 Ω resistor, a 4 V battery and an 8 V battery and a 2 Ω resistor.

A common method is as follows:

• Denote the potential difference between P and Q by V.

• In the battery branch the net drop is

  V = (net emf) – (current × total resistance)

  Here the net emf (taken in the proper sign‐convention) comes out as 4 V – 8 V = –4 V. (One may choose the sign so that eventually the answer is positive.)

• Thus writing

  V = 4 – 8 – (1+2) I ⟹ V = –4 – 3I  (1)

• In the 9 Ω branch, by Ohm’s law, V = 9 I  (2)

Equate (1) and (2):

9 I = –4 – 3I ⟹ 12 I = –4 ⟹ I = –1/3 A.

The minus sign means that our assumed direction was opposite to the actual one. In magnitude the answer is

I = 1/3 A and the potential difference (magnitude) is V = 9×(1/3) = 3 V.

Thus one acceptable answer is:

Current = 1/3 A and V₍PQ₎ = 3 V.