Question

Two batteries of emf 4V and 8V with internal resistance 1Ω and 2Ω are connected in a circuit with resistance of 9Ω as shown in figure. The current and potential difference between the points P and Q are -

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

Answer 1

Answer: (A)

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

Answer 2

i = $\frac{4}{9 + 3} = \frac{1}{3}A$ ⇒ V_P – V_Q = $\frac{1}{3}$ × 9 = 3 volt