Question

In an R-L-C series circuit shown in the figure, the readings of voltmeters $V_1$ and $V_2$ are 100 $V$ and 120 $V$ respectively. The source voltage is 130 $V$ . For this situation, mark out the correct statement(s).

• A. Voltage across resistor, inductor and capacitor are $50V,50\sqrt{3}V$ and $(120+50\sqrt{3})V$ respectively.
• B. Voltage across resistor, inductor and capacitor are $50V,50\sqrt{3}V$ and $120-50\sqrt{3}V$ respectively.
• C. Power factor of the circuit is $\frac{5}{13}$.
• D. The circuit is capacitive in nature.

Answer 1

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

The circuit is capacitive in nature.

In R-L-C series circuit,
$V_R^2+V_L^2={100}^2$
$\left(V_c-V_L\right|=120$

$${130}^2=V_R^2+{(V_C-V_L)}^2\Rightarrow V_R=50V$$
$\Rightarrow$ Voltage across inductor, $V_L=\sqrt{{100}^2-{50}^2}=50\sqrt{3}V$
and voltage drop across capacitor,
$V_C=(120+50\sqrt{3}V)$
Power factor, $\cos \varphi =\frac{V_R}{V_Z}=\frac{50}{130}=\frac{5}{13}$
Since, $V_C>V_L$, so circuit is capacitive.