Question

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

• A. $1V$
• B. $2V$
• C. $3V$
• D. $4V$

Answer 1

Answer: (D)

$4V$

Let the current across the resistors $6\Omega$ and $R\Omega$ in series be $I$.
Then,
$I\times 6=3V\Rightarrow I=0.5A$
also,
$I\times (6+R)=5V\Rightarrow R=4\Omega$
When $R$ replaced by $L$, such that current remains same,
Let impedance of inductor be $X_L$, then as current remains same,
$I=0.5A=\frac{V}{\sqrt{R^2+X_L^2}}=\frac{5}{\sqrt{6^2+X_L^2}}\Rightarrow X_L=8\Omega$
Therefore,
P.d across $L$ = $I\times X_L=0.5\times 8=4V$