Power in AC Circuits: Practice Problem & Solution
The potential differences across the resistance, capacitance and inductance are $80 V$, $40 V$ and $100 V$ respectively in an LCR circuit. The power factor of this circuit is: (2016 - II)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Power in AC Circuits. Understanding the underlying formula is key to arriving at the correct answer below:
The applied voltage is $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Substituting values, $V = \sqrt{80^2 + (100 - 40)^2} = \sqrt{6400 + 3600} = 100 V$.
The power factor is $\cos\phi = \frac{V_R}{V}$.
$\cos\phi = \frac{80}{100} = 0.8$.
Detailed Solution (Best of Understanding Complex Concepts)
Concepts Involved
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Series LCR Circuit Voltages: In a series LCR (Inductor-Capacitor-Resistor) circuit, the current $I$ is the same throughout all components. However, the voltages across them have different phase relationships with respect to the current:
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Voltage across the resistor ($V_R$) is in phase with the current.
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Voltage across the inductor ($V_L$) leads the current by $90^\circ$.
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Voltage across the capacitor ($V_C$) lags the current by $90^\circ$.
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Net Voltage (Phasor Addition): Because $V_L$ and $V_C$ are completely out of phase ($180^\circ$ apart), they partially cancel each other out. The net reactive voltage is $(V_L - V_C)$. The total applied RMS voltage $V$ is the vector sum of $V_R$ and the net reactive voltage, calculated using the Pythagorean theorem:$$V = \sqrt{V_R^2 + (V_L - V_C)^2}$$
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Power Factor ($\cos\phi$): The power factor is defined as the cosine of the phase angle ($\phi$) between the total voltage and the current. It represents the ratio of the real power dissipated (in the resistor) to the apparent power supplied. It can be calculated using the voltage values:$$\cos\phi = \frac{V_R}{V}$$
Step-by-Step Solution
Step 1: Identify the given values.
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Voltage across the resistance, $V_R = 80 \text{ V}$
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Voltage across the inductance, $V_L = 100 \text{ V}$
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Voltage across the capacitance, $V_C = 40 \text{ V}$
Step 2: Calculate the total applied voltage ($V$).
Using the phasor addition formula for a series LCR circuit:
$$V = \sqrt{V_R^2 + (V_L - V_C)^2}$$
Substitute the given values into the equation:
$$V = \sqrt{80^2 + (100 - 40)^2}$$
$$V = \sqrt{80^2 + (60)^2}$$
$$V = \sqrt{6400 + 3600}$$
$$V = \sqrt{10000}$$
$$V = 100 \text{ V}$$
Step 3: Calculate the power factor.
Use the power factor formula relating the resistive voltage to the total voltage:
$$\text{Power Factor} = \cos\phi = \frac{V_R}{V}$$
Substitute the values of $V_R$ and the calculated $V$:
$$\cos\phi = \frac{80}{100}$$
$$\cos\phi = 0.8$$
The power factor of the circuit is $0.8$ (Option 1).
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