Question

For the series LCR circuit connected with $220\,\text{V},\,50\,\text{Hz}$ a.c. source as shown in the figure, the power factor is $\dfrac{\alpha}{10}$. The value of $\alpha$ is ___.

• A. $4$
• B. $8$
• C. $10$
• D. $6$
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Hint:

In a series LCR circuit, the power factor depends only on resistance and total impedance, not on the applied voltage.

Answer 1

The Correct Option is D

Step 1: Identifying the given quantities from the circuit.
From the given LCR circuit:
Inductive reactance, $X_L = 70\,\Omega$
Capacitive reactance, $X_C = 150\,\Omega$
Resistance, $R = 60\,\Omega$
Step 2: Calculating the net reactance of the circuit.
For a series LCR circuit, net reactance is given by:
\[ X = X_L - X_C \] \[ X = 70 - 150 = -80\,\Omega \] The negative sign shows that the circuit is capacitive in nature.
Step 3: Calculating the impedance of the circuit.
The impedance $Z$ of a series LCR circuit is:
\[ Z = \sqrt{R^2 + X^2} \] \[ Z = \sqrt{(60)^2 + (80)^2} \] \[ Z = \sqrt{3600 + 6400} \] \[ Z = \sqrt{10000} = 100\,\Omega \] Step 4: Calculating the power factor.
Power factor is defined as:
\[ \cos\phi = \dfrac{R}{Z} \] \[ \cos\phi = \dfrac{60}{100} = 0.6 \] Step 5: Finding the value of $\alpha$.
Given that the power factor is $\dfrac{\alpha}{10}$,
\[ \dfrac{\alpha}{10} = 0.6 \] \[ \alpha = 6 \] Step 6: Final conclusion.
Hence, the value of $\alpha$ is $6$, which corresponds to Option (4).