GATE 2018 EE Q.48: solution and answer

GATE 2018 · EEQ.48 · 2 marks · NAT

The voltage across the circuit in the figure, and the current through it, are given by the following expressions:

$v (t) = 5 - 10 cos (ω t + 6 0^{\circ}) V$

$i (t) = 5 + X cos (ω t) A$

where $ω = 100 π$ radian/s. If the average power delivered to the circuit is zero, then the value of $X$ (in Ampere) is ______ (up to 2 decimal places).

GATE 2018 EE question 48: A two-terminal box labelled "Electrical Circuit". — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

ANSWER

$X = 10.00 A$ : the circuit absorbs $25 W$ at DC and delivers $25 W$ back at $50 Hz$ , so the net average power is zero.

Official answer: 10.0Source: Official answer key, GATE 2018 EE (IIT Guwahati)

Passive sign convention. In the figure $i (t)$ enters the $+$ terminal of $v (t)$ , so the average power delivered to the circuit is the time average of $v (t) i (t)$ : $P = \overline{v (t) i (t)}$ .
Power adds across different frequencies. Multiplying out $v (t) i (t)$ gives a DC×DC term, two DC×AC cross terms and one AC×AC term. Each cross term is a constant times a sinusoid at $ω$ , whose average over a period is zero. So $P = P_{DC} + P_{ω}$ .
DC part: $P_{DC} = V_{0} I_{0} = 5 \times 5 = 25 W$ .
Write the AC voltage as a cosine with a positive amplitude: $- 10 cos (ω t + 6 0^{\circ}) = 10 cos (ω t + 6 0^{\circ} - 18 0^{\circ}) = 10 cos (ω t - 12 0^{\circ}) V$ . So $θ_{v} = - 12 0^{\circ}$ and $θ_{i} = 0^{\circ}$ .
AC part, using peak amplitudes ( $V_{m} = 10 V$ , $I_{m} = X A$ ): $P_{ω} = \frac{V _{m} I _{m}}{2} cos (θ_{v} - θ_{i}) = \frac{10 X}{2} cos (- 12 0^{\circ}) = 5 X (- \frac{1}{2}) = - 2.5 X W$ (Same result with rms values: $\frac{10}{2} \cdot \frac{X}{2} cos 12 0^{\circ} = - 2.5 X W$ .)
Set the average power to zero: $25 - 2.5 X = 0 \Rightarrow X = \frac{25}{2.5} = 10.00 A$ .
Check: with $X = 10$ , the circuit absorbs $25 W$ at DC and $P_{ω} = - 2.5 \times 10 = - 25 W$ at $ω = 100 π rad/s$ ( $50 Hz$ ), meaning it sends $25 W$ back. The net average power is $0$ .

KEY IDEA

When voltage and current contain components at different frequencies (here DC and $ω$ ), the average power is the sum of the powers at each frequency, because products of components at different frequencies average to zero. At DC, $P = V_{0} I_{0}$ . At $ω$ , $P = \frac{1}{2} V_{m} I_{m} cos (θ_{v} - θ_{i})$ , and only after both signals are written as cosines with positive amplitudes.

WATCH OUT FOR

Taking the voltage angle as $+ 6 0^{\circ}$ and ignoring the minus sign in $- 10 cos (ω t + 6 0^{\circ})$ . That gives $P_{ω} = + 2.5 X$ and $X = - 10$ , a sign error. The minus sign is a $18 0^{\circ}$ phase shift, so the angle difference is $12 0^{\circ}$ , not $6 0^{\circ}$ . A second mistake is to leave out the DC power $5 \times 5 = 25 W$ , or to add cross terms such as $5 \cdot X cos ω t$ , which average to zero.

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 34 AC, power & resonance questions, with notes

GATE 2018 EE Q.48: solution and answer

Answer and solution

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