GATE 2022 EE Q.58: solution and answer

GATE 2022 · EEQ.58 · 2 marks · NAT

Consider an ideal full-bridge single-phase DC-AC inverter with a DC bus voltage magnitude of 1000 V. The inverter output voltage $v (t)$ shown below, is obtained when diagonal switches of the inverter are switched with 50 % duty cycle. The inverter feeds a load with a sinusoidal current given by, $i (t) = 10 sin (ω t - \frac{π}{3})$ A, where $ω = \frac{2 π}{T}$ . The active power, in watts, delivered to the load is ______. (round off to nearest integer)

GATE 2022 EE question 58: Plot of v(t) against t (sec). — Figure from the GATE 2022 EE paper. It is part of the question.

Answer and solution

ANSWER

The active power delivered to the load is about 3183 W.

Official answer: 3170 to 3190Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

The output $v (t)$ is a square wave of amplitude $V_{d c} = 1000 V$ : it is $+ 1000 V$ for $0 < t < 0.5 T$ and $- 1000 V$ for $0.5 T < t < T$ . The diagonal switches conduct for half a cycle each, so the duty ratio is $D = 0.5$ . No firing angle applies, so $α = 0$ is taken as the reference. Over one cycle the waveform has odd symmetry, so it has zero average value and contains only odd sine harmonics.
Fourier series of this square wave: $v (t) = \frac{4 V _{d c}}{π} (sin ω t + \frac{1}{3} sin 3 ω t + \frac{1}{5} sin 5 ω t + \dots)$
Fundamental peak voltage: $V_{1 m} = \frac{4 V _{d c}}{π} = \frac{4 \times 1000}{π} \approx 1273.24 V$ Its phase is $0^{\circ}$ , because the term is $sin ω t$ .
The load current is a pure sinusoid at the fundamental frequency: $I_{m} = 10 A$ , phase $- π /3$ . Average power is the sum, over harmonics, of $\frac{1}{2} V_{k m} I_{k m} cos ϕ_{k}$ . Every harmonic with $k \geq 3$ has zero current, so it contributes zero power. Only the fundamental delivers power.
Phase angle between the fundamental voltage and the current: $ϕ = 0 - (- \frac{π}{3}) = \frac{π}{3} = 6 0^{\circ}, cos ϕ = 0.5$
Active power: $P = \frac{V _{1 m} I _{m}}{2} cos ϕ = \frac{1273.24 \times 10}{2} \times 0.5 \approx 3183.1 W$
Rounded to the nearest integer, $P \approx 3183 W$ .

KEY IDEA

With a sinusoidal load current, only the fundamental component of the inverter's square-wave voltage delivers average power. Use $P = \frac{V _{1 m} I _{m}}{2} cos ϕ$ with $V_{1 m} = \frac{4 V _{d c}}{π}$ , where $ϕ$ is the phase of the fundamental voltage relative to the current.

WATCH OUT FOR

Using the full rms of the square wave ( $1000 V$ ) with the rms current, which gives $\frac{1000 \times 10}{2 \cdot 2} \times 0.5 = 2500 W$ instead of $3183 W$ . The harmonic voltages carry no power with a sinusoidal current, so the fundamental rms ( $\frac{4 \times 1000}{π 2}$ ) is the one to use. Another slip is taking the phase from the cosine reference instead of the sine reference, which changes $cos ϕ$ .

From the GATE 2022 EE paper. Every question of this paper · Download the full paper · All 6 Inverters & PWM questions, with notes

GATE 2022 EE Q.58: solution and answer

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