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Step 1: Calculate the Inductive Reactance
Use the formula for inductive reactance
$X_L = 2 \pi f L.$
Given:
• Frequency $f = 50\ \text{Hz}$
• Inductance $L = 20\ \text{mH} = 20 \times 10^{-3}\ \text{H}$
Thus,
$X_L = 2 \pi \times 50 \times (20 \times 10^{-3}) = 2 \pi \times 50 \times 0.02 = 2 \pi .$
Numerically, $2 \pi \approx 6.283\ \Omega. $
Step 2: Calculate the Capacitive Reactance
Use the formula for capacitive reactance
$X_C = \dfrac{1}{2 \pi f C}.$
Given:
• Frequency $f=50\ \text{Hz}$
• Capacitance $C = 120\,\mu\text{F}=120\times10^{-6}\ \text{F}$
So,
$X_C = \dfrac{1}{2 \pi \times 50 \times 120 \times 10^{-6}}.$
Calculate the denominator first:
$2 \pi \times 50 \times 120 \times 10^{-6} \approx 0.0377,$
hence,
$X_C = \dfrac{1}{0.0377} \approx 26.53\ \Omega.$
Step 3: Find the Net Reactance
In a series RLC circuit, the net reactance $X_{\text{net}} = X_L - X_C.$
Here,
$X_{\text{net}} = 6.283 - 26.53 = -20.247\ \Omega.$
We take the magnitude when computing impedance.
Step 4: Compute the Impedance of the Circuit
The impedance $Z$ of the series RLC circuit is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}.$
Given the resistor $R = 60\ \Omega,$ we have
$Z = \sqrt{(60)^2 + (-20.247)^2} \approx \sqrt{3600 + 410} = \sqrt{4010} \approx 63.36\ \Omega.$
Step 5: Determine the RMS Current
The series circuit is driven by a 24 V (rms) AC source. The rms current $I$ is given by
$I = \dfrac{V_{\text{rms}}}{Z} = \dfrac{24}{63.36} \approx 0.379\ \text{A}.$
Step 6: Calculate the Power Dissipated
Power dissipation in the series RLC circuit occurs only through the resistor. The average power $P$ is:
$P = I^2 R = (0.379)^2 \times 60 \approx 8.66\ \text{W}.$
Step 7: Find the Energy Dissipated in 60 s
Energy dissipated $E = P \times t = 8.66\times 60 \approx 520\ \text{J}.$
In scientific notation, this is about $5.17 \times 10^2\ \text{J},$ which matches the correct answer.
