Question

Two resistors are connected in a series across $5\text{V}$ $rms$ source of alternating potential. The potential difference across $6\Omega$ resistor is $3\text{V}$. If $R$ is replaced by a pure inductor $L$ of such magnitude that current remains same, then the potential difference across $L$ is

• A. $1\text{V}$
• B. $2\text{V}$
• C. $3\text{V}$
• D. $4\text{V}$

Answer 1

The correct option is D $4\text{V}$

In the initial case :

Current in the circuit,

$\Rightarrow I_1=\frac{5}{6+R}$

Potential difference across $6\Omega$ resistor

$\Rightarrow I_1\times 6=3$

$\Rightarrow \frac{30}{6+R}=3$

$\Rightarrow R=4\Omega$

Hence, current is $I_1=0.5\text{A}$

In, the second case:

$Z=\sqrt{36+X_L^2}$

As, the magnitude of current also remains same in this case .

$\therefore I_2=0.5\text{A}$

Also, $I_2=\frac{V}{Z}$

$\Rightarrow \frac{1}{2}=\frac{5}{\sqrt{36+X_L^2}}$

$\Rightarrow X_L=8\Omega$

Now, potential difference across $L$$=I_2{X}_L=\frac{1}{2}\times 8=4\text{V}$

Hence, option $\left(\text{D}\right)$ is correct.

Alternate solution: Given: $V=5\text{V};V_R=3\text{V}$ $V_L=?$ $\therefore V_L=\sqrt{V^2-V_R^2}=4\text{V}$