Test: Average Power - Electrical Engineering (EE) MCQ

The expression of the average power delivered to the circuit is P_avg = I_m² R/2. Given I_m = 5, R = 5. So the average power delivered to the circuit = 5²×5/2 = 62.5W.

The expression of effective value of current is I_eff = I_m/√2. Given I_m = 15. On substituting the value in the equation we get effective value of current = 15/√2 = 11V.

If a circuit has complex impedance, the average power is power dissipated in resistor only and is not stored in capacitor or inductor.

In case of purely capacitive circuit, the phase angle between the voltage and current is zero that is θ = 90⁰. Hence the average power = 0.

To get average power we have to take the product of the effective values of both voltage and current multiplied by cosine of the phase angle between the voltage and current. The expression of average power is Average power (P_avg) = V_effI_effcosθ

The expression of average power delivered to the circuit is P_avg = V_effI_effcosθ, θ = 30⁰. We have V_eff = 71, I_eff = 11. So the average power delivered to the circuit P_avg = 71 x 11 x cos 30⁰ = 650W.

The expression of effective value of voltage is V_eff = V_m/√2. Given V_m = 100. On substituting the value in the equation we get effective value of voltage = 100/√2 = 71V.

In case of purely inductive circuit, the phase angle between the voltage and current is zero that is θ = 90⁰. Hence the average power = 0.

In case of purely resistive circuit, the phase angle between the voltage and current is zero that is θ=0⁰. Hence the average power = V_mI_m/2.

To find the average value of any power function we have to take a particular time interval from t1 to t2, by integrating the function we get the average power. The equation of the average power (P_avg) is P_avg = (V_mI_m/2)cosθ.