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A $10 Ω$ resistor is carrying a current $i = (3 + 4 \sqrt{2} \sin 100 πt) A$ then potential difference between the ends of the resistor measured by an a.c. ammeter is

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detailed solution

Correct option is C

Irms=32+4222A=5A∴ Potential difference =Irms.R=5×10 volt=50 volt

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A 10 Ω resistor is carrying a current i=3+42sin100πtA then potential difference between the ends of the resistor measured by an a.c. ammeter is

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Ready to Test Your Skills?

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A $10 Ω$ resistor is carrying a current $i = (3 + 4 \sqrt{2} \sin 100 πt) A$ then potential difference between the ends of the resistor measured by an a.c. ammeter is