© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the given parameters
• Inductor, L = 20 mH = 20 × 10⁻³ H
• Capacitor, C = 100 μF = 100 × 10⁻⁶ F
• Resistor, R = 50 Ω
• Source of emf: V(t) = 10 sin(314 t)
The angular frequency of the source is ω = 314 rad/s, and the amplitude of voltage is Vm = 10 V.
Step 2: Calculate inductive and capacitive reactances
The inductive reactance is given by
$X_L = \omega L$
Substituting ω = 314 rad/s and L = 20 × 10⁻³ H:
$$
X_L = 314 \times (20 \times 10^{-3}) = 314 \times 0.02 = 6.28\ \Omega
$$
The capacitive reactance is given by
$X_C = \frac{1}{\omega C}$
Substituting ω = 314 rad/s and C = 100 × 10⁻⁶ F:
$$
X_C = \frac{1}{314 \times (100 \times 10^{-6})} = \frac{1}{314 \times 10^{-4}} = \frac{1}{0.0314} \approx 31.85\ \Omega
$$
Step 3: Calculate the net impedance Z of the series circuit
Since the resistor, inductor, and capacitor are in series, the total impedance is
$$
Z = \sqrt{R^2 + (X_C - X_L)^2}.
$$
Here,
$$
(X_C - X_L) = 31.85 - 6.28 = 25.57\ \Omega.
$$
Therefore,
$$
Z = \sqrt{(50)^2 + (25.57)^2} = \sqrt{2500 + 654.8249} \approx \sqrt{3154.8249} \approx 56.15\ \Omega.
$$
Step 4: Compute the RMS current through the circuit
The RMS value of the supply voltage is
$V_{\mathrm{rms}} = \frac{V_m}{\sqrt{2}} = \frac{10}{\sqrt{2}} \ \text{V}.$
Then the current in the circuit is
$$
I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{Z} = \frac{\frac{10}{\sqrt{2}}}{56.15} \approx \frac{7.07}{56.15} \approx 0.126 \ \text{A}.
$$
Step 5: Calculate the average power loss in the circuit
In an LCR series circuit, the average power loss is due to the resistor only and is given by
$$
P = I_{\mathrm{rms}}^2 \, R.
$$
Substituting the values:
$$
P = (0.126)^2 \times 50 \approx 0.79 \ \text{W}.
$$
Final Answer
The power loss in the circuit is 0.79 W.
