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

Answer: 600

Current with both switches opened, $\frac{V}{R_{eq}}=\frac{1.5}{450}=\frac{1}{300}=i$
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, $i_2=\frac{7}{1800}-\frac{1}{300}=\frac{1}{1800};i_{2}R=i_1{R}_1$
$\Rightarrow R=180\left(\frac{1}{3}\right)=600\Omega$