Question

In a $R-L-C$ series circuit shown, the readings of voltmeters $V_1$ and $V_2$ are $100V$ and $120V$. Choose the correct statement$\left(s\right)$.

• A. Voltage across resistor, inductor and capacitor are $50V,86.6V$ and $206.6V$ respectively.
• B. Voltage across resistor, inductor and capacitor are $10V,90V$ and $30V$ respectively.
• C. Power factor of the circuit is $\frac{5}{13}$.
• D. Circuit is capacitive in nature.

Answer 1

Answer: (A), (C), (D)

Voltage across resistor, inductor and capacitor are $50V,86.6V$ and $206.6V$ respectively.
Power factor of the circuit is $\frac{5}{13}$.
Circuit is capacitive in nature.

$\sqrt{V_R^2+V_L^2}=100$ .........$\left(i\right)$
$V_L\sim V_C=120$ .............$\left(ii\right)$
$\sqrt{V_R^2+(V_L\sim V_C{)}^2}=130$ ...........$\left(iii\right)$
Solving these three equations we get,
$V_R=50V,V_L=86.6V$ and $V_C=206.6V$
Power factor $=\cos \varphi =\frac{R}{Z}=\frac{V_R}{V}=\frac{50}{130}=\frac{5}{13}$
Since $V_C>V_L$, circuit is capacitive in nature.