An electric circuit is given below. V_{1}=1 V and Resistance R=1000 \Omega. The current through the resistance R is very close to 1 mA and the voltage across point A and B, V_{AB}=1 V. Now the circuit is changed to: where value of V_{2}=5 V. The internal resistances of both the batteries are 0.1 \Omega. The current through the resistance R is about:
- A. 1.0 mA
- B. 1.2 mA
- C. 3.0 mA ✓
- D. 5.0 mA
Correct Answer: C. 3.0 mA
Explanation
When two batteries are connected in parallel with equal internal resistance, the equivalent voltage is the average of their EMFs if polarities match appropriately, yielding 3V. The current through a 1000 \Omega resistor would be approximately 3.0 mA.
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