Question

A series LCR circuit containing a resistor of $120\Omega$ has an angular resonant frequency $4\times {10}^{5}rad{s}^{-1}$. At resonance the voltages across resistor and inductor are 60 V and $40\text{V}$ respectively.
The value of inductance L is

• A. $0.1\text{mH}$
• B. $0.2\text{mH}$
• C. $0.35\text{mH}$
• D. $0.4\text{mH}$

Answer 1

The correct option is B $0.2\text{mH}$

Given that
$\omega =4\times {10}^5\text{r}ad/s$
$V_R=60\text{V}$
$V_L=40\text{V}$
We know that
$V_L=i_{rms}{X}_L$ .......(i)

Now the circuit is given at resonance
So $X_L=X_C$
and the circuit will behave as purely resistive.
$\Rightarrow Z=R$
$\therefore I_{rms}=\frac{V_{rms}}{Z}=\frac{V_{rms}}{R}\Rightarrow I_{rms}=\frac{60}{120}=0.5\text{A}$
Putting this in eq(i)
$\Rightarrow 40=0.5\times 4\times {10}^5\times L\Rightarrow L=0.2\text{m}H$