Question

In series $LCR$ circuit voltage drop across resistance is $8V$, across inductor is $6V$ and across capacitor is $12V$. 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 $4/5$
• D. None of these

Answer 1

The correct option is C Power factor of circuit will be $4/5$

correct answer : option D
taking $I_0$ as reference ,
voltage across $R=I_{0}R=8V$
voltage across $L=j{I}_0{X}_L=j6V$
voltage across $C=-j{I}_0{X}_C=j12V$
total voltage across

$LCR=I_{0}R+j{I}_0{X}_L-j{I}_0{x}_C$

$=I_0(R+j(X_L-X_C\left)\right)$

$=8+j6-j12=8-j6=10\angle (-ta{n}^{-1}(6/8\left)\right)V$

we see that voltage across LCR lags behind the current.
we see that voltage across each element is not less than the applied voltage ( $V_C=12V$).
power factor=$\frac{R}{Z}=\frac{R}{\sqrt{R^2+(X_L-X_C{)}^2}}$

$=\frac{I_{0}R}{I_0\sqrt{R^2+(X_L-X_C{)}^2}}=\frac{I_{0}R}{\sqrt{I_0^2{R}^2+I_0^2(X_L-X_C{)}^2}}$

$=\frac{I_{0}R}{\sqrt{I_0^2{R}^2+(I_0{X}_L-I_0{X}_C{)}^2}}$

$=\frac{V_R}{\sqrt{V_R^2+(V_L-V_C{)}^2}}=\frac{8}{\sqrt{8^2+(6-12{)}^2}}\frac{8}{10}=\frac{4}{5}$