Question

Two batteries of e.m.f 4 V and 8 V with internal resistance 1$\Omega$ and 2$\Omega$ respectively are connected in series (opposing) with a 9$\Omega$ resistor. The current and potential difference between points 'P' and 'Q' is \dots

• A. $1/3$ A and 4 V
• B. $1/3$ A and 3 V
• C. $1/2$ A and 5 V
• D. $1/6$ A and 3 V

Hint:

Always remember that while opposing batteries subtract their voltages, their internal resistances always add to the total circuit resistance. Current must still force its way physically through both batteries!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When batteries are connected in opposition, the net e.m.f. is the difference between them. Total resistance is the sum of all internal and external resistances.

Step 2: Formula Application:
Net $E = 8V - 4V = 4V$. Total $R = 1\Omega + 2\Omega + 9\Omega = 12\Omega$. Current $I = E / R = 4 / 12 = 1/3$ A.

Step 3: Explanation:
The potential difference across the 9$\Omega$ resistor (between P and Q) is $V = I \times R_{ext}$. $V = (1/3) \times 9 = 3$ V.

Step 4: Final Answer:
The current is 1/3 A and the potential difference is 3 V.