Question

In an LCR series circuit, if $\frac{1}{\omega C}<\omega L$, then the incorrect option among the following is ($\omega$-angular frequency, C-capacitance and L-inductance)

• A. Phase angle is negative
• B. Power factor is zero
• C. The circuit is predominantly inductive
• D. Current in the circuit lags source voltage

Hint:

Always check for the presence of Resistance. If R exists, Power Factor ($\cos \phi$) $\neq 0$.

Answer 1

The Correct Option is B

Step 1: Analyzing the Condition:
Given $\frac{1}{\omega C}<\omega L$, which means $X_C<X_L$. The net reactance $X = X_L - X_C$ is positive. This implies the circuit is Inductive.
Step 2: Evaluating Options:

• (C) The circuit is predominantly inductive: True, since $X_L>X_C$.
• (D) Current in the circuit lags source voltage: True. In an inductive circuit, voltage leads current (or current lags voltage).
• (A) Phase angle is negative: The phase angle $\phi$ is usually given by $\tan \phi = \frac{X_L - X_C}{R}$. Since $X_L>X_C$, $\tan \phi$ is positive in the impedance triangle convention. However, if the phase of current is expressed as $-\phi$ relative to voltage, it is often referred to as a "lagging" phase angle. Whether this specific statement is strictly "true" or "false" depends on sign convention, but it describes the lag.
• (B) Power factor is zero: The power factor is $\cos \phi = \frac{R}{Z}$. In an LCR series circuit, a resistor $R$ is present (unless specified otherwise). Therefore, $R \neq 0$ and $Z \neq 0$, so the power factor cannot be zero. A power factor of zero implies a purely reactive circuit (pure inductor or capacitor), which contradicts the "LCR series circuit" description. Thus, this statement is definitely incorrect.

Step 3: Conclusion:
Since the question asks for the incorrect option, (B) is the answer.