WAEC - Physics (1997 - No. 36)
An alternating current with a peak value of 5A passes through a resistor of resistance 10Ω. Calculate the rate at which energy is dissipated in the resistor
250W
125W
50W
35.4W
12.5W
Explanation
Given: Peak current, \( I_{\text{peak}} = 5 \, \text{A} \), Resistance, \( R = 10 \, \Omega \)
Calculate the rms current: The rms current is given by:
\(I_{\text{rms}} = \frac{I_{\text{peak}}}{\sqrt{2}} = \frac{5}{\sqrt{2}} \approx 3.54 \, \text{A}\)
Calculate the power dissipated in the resistor: Using the rms value, the power can be calculated as:
\(P = I_{\text{rms}}^2 \cdot R\)
Substituting the values:
\(P = \left(\frac{5}{\sqrt{2}}\right)^2 \cdot 10\)
Calculating:
\(P = \frac{25}{2} \cdot 10 = 125 \, \text{W}\).
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