In the circuit below, the battery \( E_1 \) has an emf of 12 V and zero internal resistance, while the battery \( E_2 \) 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
The correct answer is Option (2) → 100
When the galvanometer reads zero, no current flows in the branch with G and \( E_2 \).
So the same current flows from \( E_1 \) through the \( 500\,\Omega \) resistor and X.
From the figure, the G and \( E_2 \) branch is joined across X, so the potential difference across X equals the emf of \( E_2 \), 2 V.
The current through the \( 500\,\Omega \) resistor and X is:
\( \displaystyle I=\frac{12\,\mathrm{V}}{(500+X)\,\Omega} \)
Putting the potential difference across X equal to 2 V:
\( \displaystyle \frac{12X}{500+X}=2 \)
\( 12X=1000+2X \)
\( 10X=1000 \)
\( \Rightarrow\; X=100 \)