Question

In series $L C R$ circuit voltage drop across resistance is $8 \,V$, across inductor is $6 \,V$ and across capacitor is $12 \,V$. Then

• A. voltage of the source will be leading current in the circuit
• B. voltage drop across each element will be less than the applied voltage
• C. power factor of circuit will be $4 / 3$
• D. none of these

Answer 1

Answer: (D)

Since $\cos \phi=\frac{R}{Z}=\frac{I R}{I Z}=\frac{8}{10}=\frac{4}{5}$
(Also $\cos \phi$ can never be greater than $1$)
Hence, (c) is wrong.
Also $I X_{ C }>I X_{ L } \Rightarrow X_{ C }>X_{ L }$
$\therefore $ Current will be leading.
In an $L C R$ circuit,
$V=\sqrt{\left(V_{ L }-V_{ C }\right)^{2}+V_{ R }^{2}}=\sqrt{(6-12)^{2}+8^{2}}$
$V=10$; which is less than voltage drop across capacitor.