GATE EE 2018 Set 1 — Question 58

The average power $P_{avg}$ delivered to a circuit with periodic voltage $v(t)$ and current $i(t)$ is the sum of the average power of the DC components and the average power of the AC components at each harmonic frequency.The given voltage and current are:
$v(t) = 5 – 10 \cos(\omega t + 60^\circ) V$
$i(t) = 5 + X \cos(\omega t) A$We can separate the DC and AC components:
DC components: $V_{dc} = 5$ V, $I_{dc} = 5$ A.
AC components: $v_{ac}(t) = -10 \cos(\omega t + 60^\circ)$ V, $i_{ac}(t) = X \cos(\omega t)$ A.The average power from the DC components is:
$P_{dc} = V_{dc} \times I_{dc} = 5 \times 5 = 25$ W.The average power from the AC components is given by $P_{ac} = \frac{V_m I_m}{2} \cos(\theta_v - \theta_i)$, where $V_m$ and $I_m$ are the peak amplitudes and $\theta_v$ and $\theta_i$ are the phase angles.First, let's express the AC voltage in a standard cosine form:
$v_{ac}(t) = -10 \cos(\omega t + 60^\circ) = 10 \cos(\omega t + 60^\circ - 180^\circ) = 10 \cos(\omega t - 120^\circ)$ V.
So, the peak voltage amplitude is $V_m = 10$ V and its phase is $\theta_v = -120^\circ$.For the AC current:
$i_{ac}(t) = X \cos(\omega t)$.
The peak current amplitude is $I_m = X$ A and its phase is $\theta_i = 0^\circ$.Now, calculate the AC power:
$P_{ac} = \frac{10 \cdot X}{2} \cos(-120^\circ - 0^\circ) = 5X \cos(-120^\circ)$.
Since $\cos(-120^\circ) = -0.5$,
$P_{ac} = 5X(-0.5) = -2.5X$ W.The total average power is the sum of DC and AC power:
$P_{avg} = P_{dc} + P_{ac} = 25 - 2.5X$.The problem states that the average power delivered to the circuit is zero, so $P_{avg} = 0$.
$25 - 2.5X = 0$
$2.5X = 25$
$X = \frac{25}{2.5} = 10$.The value of X is 10.00 A.