Question

Two batteries of emf 4 V and 8 V with internal resistance 1 \( \Omega \) and 2 \( \Omega \) 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 are:

• A. \( \dfrac{4}{3} \) A and 3 V
• B. \( \dfrac{1}{6} \) A and 4 V
• C. \( \dfrac{1}{9} \) A and 9 V
• D. \( \dfrac{1}{12} \) A and 12 V

Hint:

Use Kirchhoff’s law to analyze the current and potential differences in circuits with multiple batteries and resistors.

Answer 1

The Correct Option is A

Step 1: Apply Kirchhoff's law to the circuit. The total emf is \( 4 + 8 = 12 \, \text{V} \), and the total resistance is \( 9 + 1 + 2 = 12 \, \Omega \).
Step 2: The total current is given by: \[ I = \dfrac{12}{12} = 1 \, \text{A}. \] Step 3: The potential difference between P and Q is: \[ V = IR = 1 \times 3 = 3 \, \text{V}. \]
Final Answer: \[ \boxed{\dfrac{4}{3} \, \text{A} \, \text{and} \, 3 \, \text{V}} \]