Question

What is the potential drop between points $A$ and $C$ in the following circuit?
Resistances $ 1 \Omega $ and $ 2 \Omega $ are represent the internal resistance of the respective cells.

• A. 1.75 V
• B. 2.25 V
• C. $ \frac{5}{4} V $
• D. $ \frac{4}{5} V $

Answer 1

Answer: (B)

Emf's $ E_1 $ and $ E_2 $ are opposing each other. Since,
$ E_2 > E_1 $ ,so current will flow from right to left

Current in the circuit
$ i = \frac{Net\, \, emf}{Total \, \, Resistance} = \frac{E_2 - E_1}{R + r_1 + r_2} $
Given, $ R = 5\Omega$ , $ r_1 = 1\Omega $ , $ r_2 = 2\Omega$
$ E_1 = 2V $ and $ E_2 = 4V $
$ \therefore \, \, \, i = \frac{4 - 2}{5 + 1 + 2} = \frac{2}{8} = 0.25 A $
The potential drop between points A and C,
$ V_A - V_C = E_1 + ir_1 $
$ = 2 + 0.25 \times 1 $
$ = 2.25 V $