In the below circuit, the battery E1 has an emf of 12 V and zero internal resistance. While the battery E2 has an emf of 2 V.
If the galvanometer G reads zero, then the value of the resistance X in ohms is:
Answer & explanation
Correct answer: option 2
We have, I1 = 0
∴ VX = E2 = 2V
$\frac{V_{500 \Omega}}{V_{X}}=\frac{(12-2)}{2}=\frac{500}{X}$
∴ X = 100 Ω