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Two batteries are connected in parallel across an external resistor.…

Question

Two batteries are connected in parallel across an external resistor. Battery 1 has EMF 10 V and internal resistance 1 ohm; battery 2 has EMF 4 V and internal resistance 1 ohm; both positive terminals join the same node and drive an external resistor R = 2 ohm to the other node. Find the power dissipated in R. If it equals p/q watts in lowest terms, report p + q.

✓ Verified answer: 417checked by our engine — not a guess

Step-by-step solution

Let V be the voltage across R (terminal voltage of the parallel batteries).

Kirchhoff current law at the top node: current from each battery into the node equals current through R: (E1 - V)/r1 + (E2 - V)/r2 = V/R.

So (10 - V)/1 + (4 - V)/1 = V/2, giving 14 - 2V = V/2, i.e.
28 - 4V = V, so 5V = 28, V = 28/5.
Current in R = V/R = 14/5 A.
Power = I^2 R = (14/5)^2 * 2 = 392/25 W (lowest terms).
Hence p+q = 392+25 = 417.

Final answer417

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