Question

Calculate the current drawn from the battery by the network of resistors shown in the figure.

Answer 1

The equivalent circuit is shown in the figure.

The five resistors form a balanced Wheatstone's bridge

$\frac{R_1}{R_5}=\frac{R_4}{R_3}$

So, $R_2$ is ineffective.

The effective resistance of $R_1$ and $R_5$ in series,

$R^{\prime }=R_1+R_5=1+2=3\Omega$

The effective resistance of $R_4$ and $R_3$ in series is

$R^{\prime \prime }=R_4+R_3=2+4=6\Omega$

Therefore, equivalent resistance of network between $A$ and $B$ is

$R_{AB}=\frac{R^{\prime }{R}^{\prime \prime }}{R^{\prime }+R^{\prime \prime }}=\frac{3\times 6}{3+6}=2\Omega$

Current drawn from battery,

$I=\frac{E}{R_{AB}}=\frac{4}{2}=2A$