When an alternating current given by I = 10sin (120π)t passes through a 12Ω resistor, the power dissipated in the resistor is
To find the power dissipated in a resistor due to an alternating current (AC), we use the formula for average power:
\(P = I_{\text{rms}}^2 R\)
where:
- \( P \) is the average power,
- \( I_{\text{rms}} \) is the root mean square (RMS) value of the current,
- \( R \) is the resistance.
The current is given by: \(I(t) = 10 \sin(120\pi t)\)
The peak current is: \(I_0 = 10 \, \text{A}\)
The RMS value of a sinusoidal current is given by:
\(I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, \text{A} \approx 7.07 \, \text{A}\)
Using the resistance \( R = 12 \, \Omega \):
\(P = I_{\text{rms}}^2 R\)
Substituting the values:
\(P = (5\sqrt{2})^2 \cdot 12\)
Calculating \( (5\sqrt{2})^2 \):
\((5\sqrt{2})^2 = 25 \cdot 2 = 50\)
Now substituting back into the power formula: \(P = 50 \cdot 12 = 600 \, \text{W}\)
Thus, the power dissipated in the resistor is: \(600 \, \text{W}\)
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