When a certain resistor is connected to an ac generator with a maximum voltage o

Question

When a certain resistor is connected to an ac generator with a maximum voltage of 11 V , the average power dissipated in the resistor is 35 W .

A) What is the resistance of the resistor?

b)  What is the rms current in the circuit?

c) We know that Pav=I2rmsR, and hence it seems that reducing the resistance should reduce the average power. On the other hand, we also know that Pav=V2rms/R, which suggests that reducing Rincreases Pav. Which conclusion is correct?

1) Reducing R increases Pav

2) Reducing R reduces Pav

Explanation / Answer

Vo =11 V, Pavg = 35 W

A) Pavg = Vo^2/2R

35 =11^2/(2*R)

R = 1.73 ohms

B) Vrms = Irms*R

Vrms = Vo/(2)^0.5

11/2^0.5 = Irms*1.73

Irms = 4.5 A

C) correct option is 2)

P = I^2rms*R

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