Question

In the circuit shown in figure the reading of ideal ammeter connected as shown is same with both switches open as with both closed. Then the resistance $R$ in $\Omega$.

Answer 1

Current with both switches opened, $\frac{V}{R_{e q}}=\frac{1.5}{450}=\frac{1}{300}=i$
After closing the switch, if $V_{1}$ and $V_{2}$ are potential differences as shown in figure then we have
$V_{1}+V_{2}=V \Rightarrow \frac{1}{3}+V_{2}=\frac{3}{2}$
$\Rightarrow V_{2}=\frac{9-2}{6}=\frac{7}{6} V$

Current through battery, $i=\frac{7}{6} \times \frac{1}{300}=\frac{7}{1800} A$
By KCL, $i_{2}=\frac{7}{1800}-\frac{1}{300}=\frac{1}{1800} ; i_{2} R=i_{1} R_{1}$
$\Rightarrow R=180\left(\frac{1}{3}\right)=600\, \Omega$