Question

In the circuit shown, the diode is ideal. The potential of $A$ is higher than the potential of $B$. If the switch $S$ is $\left(i\right)$ closed$,\left(ii\right)$ open the equivalent resistance between $A$ and $B$ is

• A. $R,R$
• B. $\frac{R}{2},\frac{R}{2}$
• C. $\frac{R}{3},\frac{R}{4}$
• D. $\frac{R}{4},\frac{R}{3}$

Answer 1

Answer: (D)

$\frac{R}{4},\frac{R}{3}$

Rearranging the given circuit into its equivalent circuit as shown in figure. All the resistances are parallel to each other hence voltage across all resistances is same. When the switch is closed current will flow through all branches i.e. all the resistances are connected in the circuit. Hence, the equivalent resistance is
$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}$
$\frac{1}{R_{eq}}=\frac{4}{R}$
$R=\frac{R}{4}$
When switch is open, current will not flow through diode and hence through $GH$. Thus, resistance R of this branch is not connected in circuit, the equivalent resistance is now
$\frac{1}{R_{eq}^{\prime }}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}$
$\frac{1}{R_{eq}^{\prime }}=\frac{3}{R}$
$R^{\prime }=\frac{R}{3}$