Power in AC Circuits: Practice Problem & Solution
An inductor $20mH$, a capacitor $100 \mu F$ and a resistor $50 \Omega$ are connected in series across a source of emf, $V = 10 sin 314 t$. The power loss in the circuit is (2018)
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:
Reactances are $X_L = \omega L = 314 \times 0.020 = 6.28 \Omega$ and $X_C = \frac{1}{314 \times 100 \times 10^{-6}} \approx 31.85 \Omega$.
Impedance $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{50^2 + (25.57)^2} \approx 56.15 \Omega$.
$V_{rms} = \frac{10}{\sqrt{2}} = 5\sqrt{2} V$, so power $P = \frac{V_{rms}^2 R}{Z^2}$.
$P = \frac{50 \times 50}{(56.15)^2} \approx 0.79 W$.
Detailed Solution (Best of Understanding Complex Concepts)
Concepts Involved
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Series LCR Circuit: The behavior of an electrical circuit where a resistor (R), inductor (L), and capacitor (C) are connected in series with an alternating current (AC) voltage source.
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Reactance ($X$): The opposition to current flow in an AC circuit caused by inductors (Inductive Reactance, $X_L$) and capacitors (Capacitive Reactance, $X_C$).
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Impedance ($Z$): The total effective resistance to alternating current, combining both standard resistance and reactance.
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Average Power ($P$): In an LCR circuit, true power is only dissipated across the resistor. It is determined by the RMS voltage, RMS current, and the circuit's power factor ($\cos\phi$).
Step-by-Step Solution
Step 1: Identify the given parameters
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Inductance, $L = 20\text{ mH} = 20 \times 10^{-3}\text{ H}$
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Capacitance, $C = 100{ \mu F} = 100 \times 10^{-6}\text{ F}$
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Resistance, $R = 50\text{ }\Omega$
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Voltage equation, $V = 10 \sin(314t)$
Step 2: Extract peak voltage and angular frequency
By comparing the given equation $V = 10 \sin(314t)$ with the standard AC voltage equation $V = V_0 \sin(\omega t)$, we get:
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Peak voltage, $V_0 = 10\text{ V}$
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Angular frequency, $\omega = 314\text{ rad/s}$
From this, the Root Mean Square (RMS) voltage is:
$$V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}}\text{ V}$$
Step 3: Calculate the Reactances
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Inductive Reactance ($X_L$):$$X_L = \omega L$$$$X_L = 314 \times 20 \times 10^{-3}$$$$X_L = 6.28\text{ }\Omega$$
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Capacitive Reactance ($X_C$):$$X_C = \frac{1}{\omega C}$$$$X_C = \frac{1}{314 \times 100 \times 10^{-6}}$$$$X_C = \frac{1}{0.0314}$$$$X_C \approx 31.85\text{ }\Omega$$
Step 4: Calculate the total Impedance ($Z$)
The formula for the total impedance of a series LCR circuit is:
$$Z = \sqrt{R^2 + (X_C - X_L)^2}$$
$$Z = \sqrt{50^2 + (31.85 - 6.28)^2}$$
$$Z = \sqrt{2500 + (25.57)^2}$$
$$Z = \sqrt{2500 + 653.82}$$
$$Z = \sqrt{3153.82}$$
$$Z \approx 56.16\text{ }\Omega$$
Step 5: Calculate the average power loss
The average power dissipated in a series LCR circuit is given by:
$$P = V_{rms} I_{rms} \cos\phi$$
Since $I_{rms} = \frac{V_{rms}}{Z}$ and the power factor is $\cos\phi = \frac{R}{Z}$, we can substitute these into the power equation:
$$P = V_{rms} \times \left(\frac{V_{rms}}{Z}\right) \times \left(\frac{R}{Z}\right)$$
$$P = \frac{V_{rms}^2 R}{Z^2}$$
Now, substitute the calculated values:
$$V_{rms}^2 = \left(\frac{10}{\sqrt{2}}\right)^2 = \frac{100}{2} = 50\text{ V}^2$$
$$P = \frac{50 \times 50}{(56.16)^2}$$
$$P = \frac{2500}{3153.94}$$
$$P \approx 0.79\text{ W}$$
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