Question

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

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

Answer 1

The Correct Option is A

To solve the given problem, we need to determine the current flowing through the circuit and the potential difference between the points P and Q.

Step-by-Step Solution:

1. Identify the Circuit Elements: The circuit consists of two batteries and three resistors. The batteries have electromotive forces (emf) of 4 V and 8 V with internal resistances of 1 Ω and 2 Ω, respectively. The third resistor has a resistance of 9 Ω.
2. Set Up the Circuit Equation: We will use Kirchhoff's loop rule. Consider the loop including both batteries and resistors connected in series. The total emf in the circuit is the algebraic sum of the emfs of the batteries, and the total resistance is the sum of all resistances in the circuit.
   • Total emf \( E = 8 \, \text{V} - 4 \, \text{V} = 4 \, \text{V} \)
   • Total resistance \( R_{\text{total}} = 1 \, \Omega + 2 \, \Omega + 9 \, \Omega = 12 \, \Omega \)
3. Calculate the Current (I): Using Ohm’s Law, \( V = IR \), where V is the potential difference, I is the current, and R is the resistance.
   • \( I = \dfrac{E}{R_{\text{total}}} = \dfrac{4 \, \text{V}}{12 \, \Omega} = \dfrac{1}{3} \, \text{A} \)
4. Calculate the Potential Difference between P and Q: The potential difference across points P and Q (the 9 Ω resistor) is given by Ohm's Law: \( V = IR \).
   • \( V_{\text{PQ}} = \left(\dfrac{1}{3} \, \text{A}\right) \times (9 \, \Omega) = 3 \, \text{V} \)

Conclusion:

The current flowing through the circuit is \( \dfrac{1}{3} \, \text{A} \) and the potential difference between points P and Q is 3 V.

Correct Answer: \( \dfrac{1}{3} \, \text{A} \) and 3 V