© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Resonance Condition in LCR Circuit
In a series LCR circuit at resonance, the inductive reactance and capacitive reactance cancel each other out, and the circuit behaves as if only the resistor were present in the circuit.
Step 2: Identify Relevant Quantities
Peak voltage applied, V_{peak} = 250 \text{ V}
Resistance, R = 8 \,\Omega
Inductance, L = 24 \text{ mH} (not needed explicitly here except to note resonance occurs)
Capacitance, C = 60 \,\mu\text{F} (not needed explicitly here except to note resonance occurs)
Step 3: Convert Peak Voltage to RMS Voltage
The RMS (Root Mean Square) voltage for a sinusoidal signal is given by:
V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}}.
So in this problem:
V_{\text{rms}} = \frac{250}{\sqrt{2}} \text{ V}.
Step 4: Apply the Power Dissipation Formula at Resonance
At resonance for a series LCR circuit, the power dissipated is given by the resistor alone:
P = \frac{V_{\text{rms}}^2}{R}.
Substituting the RMS voltage and resistor values:
P = \frac{\left(\frac{250}{\sqrt{2}}\right)^2}{8}.
Simplify the expression inside:
\left(\frac{250}{\sqrt{2}}\right)^2 = \frac{250^2}{2} = \frac{62500}{2} = 31250.
Therefore,
P = \frac{31250}{8} = 3906.25 \text{ W}.
Converting to kilowatts (kW):
P = 3.90625 \text{ kW} \approx 4 \text{ kW}.
Step 5: State the Final Answer
The value of power dissipated at the resonant condition, to the nearest integer in kW, is 4.
