The galvanometer G in the circuit reads zero (no deflection). If the batteries are resistance less, the value of R would be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 100Ω
$R_{effective}=500+R$ [Connected in series]
$V=12V$
$500×\frac{12}{500+R}+R×\frac{12}{500+R}=12$
$⇒R=100Ω$