Question

In the circuit shown in figure the internal resistance of the cell is negligible. For the value of $R=\frac{40}{x}\Omega$, no current flows through the galvanometer. What is $x$?(Integer value)

Answer 1

Potentials at nodal points are as shown in the figure.



Applying $\text{KCL}$ at point $P$ we get,

$\frac{V^{\prime }-2}{40}=\frac{2-0}{40}\Rightarrow V^{\prime }=4\text{volts}$

Let the current passing through $40\Omega$ resistor be $I_1$.

$\therefore I_1=\frac{2}{40}=0.05\text{A}$

So, from the diagram given in the question, we can say that current of $0.45\text{A}$ will be flowing through the resistance $R$.

Thus, $\frac{V^{\prime }-2}{R}=0.45$

$\Rightarrow R=\frac{2}{0.45}=\frac{40}{9}\Omega$

Comparing this with the data given in the question, $x=9$

Accepted answer : $9$