The average power dissipated in a 47 ohm resistor is 8.0 W. (a) What is the peak value i_0 of the ac current in the resistor?
Question:
The average power dissipated in a {eq}47 Omega {/eq} resistor is {eq}8.0 W {/eq}.
(a) What is the peak value {eq}i_0 {/eq} of the ac current in the resistor?
Answers (1)
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Muriel April 18, 2023 в 08:38The peak value of the AC current in the resistor can be found using the formula: {eq}P_{avg} = \frac{1}{2}I_{rms}^2 R {/eq} Where: {eq}P_{avg} = 8.0 W {/eq} is the average power dissipated in the resistor {eq}R = 47 Omega {/eq} is the resistance of the resistor Solving for {eq}I_{rms} {/eq}, we get: {eq}I_{rms} = sqrt{frac{2P_{avg}}{R}} {/eq} Substituting the given values, we get: {eq}I_{rms} = sqrt{frac{2(8.0 W)}{47 Omega}} = 0.720 A {/eq} Since we know that the current is AC, we can use the relationship between RMS and peak values: {eq}I_{pk} = sqrt{2} times I_{rms} {/eq} Substituting {eq}I_{rms} {/eq}, we get: {eq}I_{pk} = sqrt{2} times 0.720 A = 1.02 A {/eq} Therefore, the peak value of the AC current in the {eq}47 Omega {/eq} resistor is {eq}1.02 A {/eq}.
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