Question

In an RLC series circuit shown in figure, the readings of voltmeters $V_1$ and $V_2$ are $100$V and $120$V, respectively. The same voltage is $130$V. For this situation, mark out the incorrect statement.

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

Answer 1

Answer: (D)

The circuit is capacitance in nature

$V_1=100V$

$V_2=120V$

$V_3=130V$

$V_R+j{V}_L=100$(equation $1$)

$V_L-V_C=120$(equation $2$)

Using KVL in loop,

$V_R+j(V_L-V_C)=130$

$V_R+j120=130$

$V_R=50V$

$V_R+j{V}_L=100$

$50+j{V}_L=100$

$V_L=\sqrt{(100{)}^2-(50{)}^2}=\sqrt{10000-2500}=\sqrt{7500}=50\sqrt{3}V$

$V_L+V_C=120$

$V_C=120-50\sqrt{3}$

$P.F=\frac{V_R}{V}=\frac{50}{130}=\frac{5}{13}$

$V_C>V_L$

Hence, circuit is capacitive in nature.