The average power dissipated in an AC circuit is given by the formula Pavg = VIcos(Φ), where V is the peak voltage, I is the peak current, and Φ is the phase difference between the voltage and the current.

In this case, the peak voltage V is 100V, the peak current I is 100mA = 0.1A, and the phase difference Φ is π/3 rad.

Substituting these values into the formula gives:

Pavg = (100V)(0.1A)cos(π/3) = 10Wcos(π/3) = 5W.

So, the average power dissipated in one cycle is 5 Watts.

Question

The average power dissipated in an AC circuit is given by the formula Pavg = VIcos(Φ), where V is the peak voltage, I is the peak current, and Φ is the phase difference between the voltage and the current.

In this case, the peak voltage V is 100V, the peak current I is 100mA = 0.1A, and the phase difference Φ is π/3 rad.

Substituting these values into the formula gives:

Pavg = (100V)(0.1A)cos(π/3) = 10Wcos(π/3) = 5W.

So, the average power dissipated in one cycle is 5 Watts.

Knowee AI · Accepted Answer

The average power dissipated in an AC circuit is given by the formula Pavg = VIcos(Φ), where V is the peak voltage, I is the peak current, and Φ is the phase difference between the voltage and the current.

In this case, the peak voltage V is 100V, the peak current I is 100mA = 0.1A, and the phase difference Φ is π/3 rad.

Substituting these values into the formula gives:

Pavg = (100V)(0.1A)cos(π/3) = 10Wcos(π/3) = 5W.

So, the average power dissipated in one cycle is 5 Watts.

In an a.c. circuit, voltage and current are given by : V=100sin(100t)V and I=100sin(100t+π3)mA respectively.The average power dissipated in one cycle is :

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