Question

In a series $\text{LCR}$ circuit, voltage drop across resistance is $8\text{V}$ , across inductor is $6\text{V}$ and across capacitor is $12\text{V}$. Then,

• A. Voltage of the source will be leading in the circuit.
• B. Voltage drop across each element will be less than the applied voltage.
• C. Power factor of the circuit will be $\frac{3}{4}$.
• D. None of the above.

Answer 1

The correct option is D None of the above.

As, we know that,

$V_{app}^2=V_R^2+(V_C-V_L{)}^2=8^2+(12-6{)}^2={10}^2$

$\Rightarrow V_{app}=10\text{V}$

$\Rightarrow V_C>V_{app}>V_R>V_L$

So, option $\left(b\right)$ is wrong.



Since, $V_C>V_L\Rightarrow V_{app}$ will lag behind the current.

So, option $\left(a\right)$ is wrong.

Power factor,

$\cos \varphi =\frac{V_R}{V_{app}}=\frac{8}{10}=0.8$

So, option $\left(c\right)$ is also wrong.


Hence, $\left(D\right)$ is the correct answer.