GATE EE · Power Electronics

Inverters – GATE EE Power ElectronicsPractice Questions & PYQs

Generate GATE-level questions on Inverters. Focus on: 1. Single-phase and Three-phase bridge inverters: 120° and 180° conduction modes. 2. Voltage Control and PWM techniques: Single-pulse, Multiple-pulse, and SPWM. 3. Harmonic analysis and Filter design principles.

33 questions · 20 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2025MCQ

The 3-phase modulating waveforms $({v}_{{a}}({t}), {v}_{{b}}({t})$ and ${v}_{{c}}({t}) )$ , used in sinusoidal PWM in a Voltage Source Inverter (VSI) are ${v}_{{a}}({t})=0.8 \sin (\omega {t}) {V}$ ${v}_{{b}}({t})=0.8 \sin \left(\omega {t}-\frac{2 \pi}{3}\right) {V}$ ${V}_{{c}}({t})=0.8 \sin \left(\omega {t}+\frac{2 \pi}{3}\right) {V}$ Where $\omega=2 \pi \times 40 {rad} / {s}$ is the fundamental frequency. The modulating waveforms are compared with a 10 kHz triangular carrier whose magnitude varies between +1 and -1 . The VSI has a DC link voltage of 600 V and feeds a star connected motor. The per phase fundamental RMS motor voltage, in volts, is closest to

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📖 Explanation

Step 1: The modulation index $m_a$ is defined as: $m_a = \frac{V_m}{V_{\text{tri}}}$ , where $V_m = 0.8$ (peak of sine) and $V_{\text{tri}} = 1$ (peak of carrier) So: $m_a = 0.8$ Step 2: In sinusoidal PWM for a 3-phase VSI, the peak value of fundamental line-to-neutral voltage is: $V_{an}^{\text{peak}} = m_a \cdot \frac{V_{dc}}{2} = 0.8 \cdot \frac{600}{2} = 240, \text{V}$ Step 3: Convert peak to RMS for a sine wave: $V_{an}^{\text{rms}} = \frac{V_{an}^{\text{peak}}}{\sqrt{2}} = \frac{240}{\sqrt{2}} \approx 169.71, \text{V}$

Q2GATE 2024NAT

A single-phase half-controlled bridge converter supplies an inductive load with ripple free load current. The triggering angle of the converter is $60^{\circ}$ . The ratio of the rms value of the fundamental component of the input current to the rms value of the total input current of the bridge is ______ (rounded off to 3 decimal places).

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📖 Explanation

$1-\phi$ Half controlled bridge rectifier $\alpha=60^{\circ}$

\begin{aligned} g & =\frac{I_{s 1}}{I_{s r}}=\frac{\frac{2 \sqrt{2}}{\pi} I_{0} \cdot \cos \frac{\alpha}{2}}{I_{0}\left(\frac{\pi-\alpha}{\pi}\right)^{1 / 2}}=\frac{2 \sqrt{2} \cdot \cos \frac{\alpha}{2}}{\sqrt{\pi \cdot(\pi-\alpha)}} \\ & =\frac{2 \sqrt{2} \cdot \cos \left(\frac{60}{2}\right)}{\sqrt{\pi \cdot\left(\pi-\frac{\pi}{3}\right)}}=\frac{3}{\pi}=0.955 \end{aligned}

Alternate solution:

\begin{aligned} 2 d & =\pi-\alpha \\ & =\pi-\frac{3}{\pi} \\ 2 d & =\frac{2 \pi}{\pi}=120^{\circ} \\ g & =\frac{3}{\pi}=0.955 \end{aligned}

Explanation Figure 1

Q3GATE 2024NAT

In the DC-DC converter shown in the figure, the current through the inductor is continuous, The switching frequency is $500 \mathrm{~Hz}$ . The voltage $\left(V_{0}\right)$ across the load is assumed to be constant and ripple free. The peak inductor current in amperes is ______(rounded off to the nearest integer).

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📖 Explanation

\begin{aligned} V_{0} & =\frac{V_{S}}{1-\alpha} \\ 40 & =\frac{20}{1-\alpha} \\ \alpha & =0.5 \\ I_{0} & =\frac{V_{0}}{R}=\frac{40}{10}=4 \mathrm{~A} \end{aligned}

Boost :

\begin{aligned} I_{L} & =\frac{I_{0}}{1-\alpha}=\frac{4}{1-0.5}=8 \mathrm{~A} \\ \Delta I_{L} & =\frac{\alpha V_{s}}{f C}=\frac{0.5 \times 20}{500 \times 2.10^{-3}}=10 \\ I_{m x} & =(i L)_{\text {peak }} \\ & =I_{L}+\frac{\Delta I_{L}}{2}=8+\frac{10}{2}=13 \mathrm{~A} \end{aligned}

Q4GATE 2024NAT

A single-phase full bridge voltage source inverter (VSI) feeds a purely inductive load. The inverter output voltage is a square wave in $180^{\circ}$ conduction mode. The fundamental frequency of the output voltage is $50 \mathrm{~Hz}$ . If the DC input voltage of the inverter is $100 \mathrm{~V}$ and the value of the load inductance is $20 \mathrm{mH}$ , the peak-to-peak load current in amperes is ______ (rounded off to the nearest integer).

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📖 Explanation

\begin{aligned} & 1-\phi \mathrm{VSI}, f_{1}=f_{0}=50 \mathrm{~Hz} \\ & V_{S}=100 \mathrm{~V} \end{aligned}

$V_{0}=V_{S} \quad\left[0 \leq t \leq \frac{T}{2}\right]$

\begin{array}{rlrl} L \cdot \frac{d i}{d t} & =V_{S} \\ \frac{d i}{d t} & =\frac{V_{S}}{L} \\ \int d i_{0} & =\frac{V_{S}}{L} \int d t \\ i_{0} & =\frac{V_{S}}{L} t+K & \\ -I_{m} & =\frac{V_{S}}{L}(0)+K & \\ i_{0} & =\frac{V_{S}}{L} \cdot t-I_{m} & \text { At } t=0, i_{0}=-I_{m} \end{array}

\text { At } t=\frac{T}{2}, \quad \begin{aligned} i_{0} & =I_{M} \\ i_{0} & =\frac{V_{S}}{L} \cdot t-I_{m} \\ I_{M} & =\frac{V_{S}}{L} \cdot \frac{I}{2}-I_{m} \\ 2 I_{m} & =\frac{V_{S}}{2 f L} \\ I_{m} & =\frac{V_{S}}{4 f L} \end{aligned}

$\therefore$ Peak to peak load current, $2 I_{m}=\frac{V_{S}}{2 f L}=\frac{100}{2.50\left(20 \cdot 10^{-3}\right)}=50 \mathrm{~A}$

Explanation Figure 1

Q5GATE 2022NAT

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, $i(t)=10 \sin (\omega t-\frac{\pi}{3})A$ , where $\omega =\frac{2\pi}{T}$ . The active power, in watts, delivered to the load is _________. (round off to nearest integer)

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📖 Explanation

For $1-\phi$ invertor, RMS value of fundamental component $V_{01}=\frac{2\sqrt{2}V_s}{\pi}$ Now, Power output $=V_{01}I_{or} \cos \frac{\pi}{3}=\frac{2\sqrt{2}}{\pi} \times 1000 \times \frac{10}{\sqrt{2}} \times \frac{1}{2}=3183.098W$

Q6GATE 2022NAT

A 3-phase grid-connected voltage source converter with DC link voltage of 1000 V is switched using sinusoidal Pulse Width Modulation (PWM) technique. If the grid phase current is 10 A and the 3-phase complex power supplied by the converter is given by $(-4000 - j3000) VA$ , then the modulation index used in sinusoidal PWM is ___________. (round off to two decimal places)

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📖 Explanation

Apparent power, S = (-4000 - j3000) VA or $S=5000\angle -143.13^{\circ}VA$ $S=\sqrt{3}V_LI_L$ $V_L=\frac{5000}{\sqrt{3} \times 10}=288.675V$ We have, Peak, $V_{01}=\sqrt{3}M_A\frac{V_s}{2}$ Put the values $\sqrt{2}\times 288.675=\sqrt{3}M_A \times \frac{1000}{2}$ $\Rightarrow \; M_A=0.471$

Q7GATE 2021NAT

A single-phase full-bridge inverter fed by a $\text{325 V DC}$ produces a symmetric quasi-square waveform across 'ab' as shown. To achieve a modulation index of 0.8, the angle $\theta$ expressed in degrees should be _______________. (Round off to 2 decimal places.) (Modulation index is defined as the ratio of the peak of the fundamental component of $V_{ab}$ to the applied $\text{DC}$ value.)

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📖 Explanation

\begin{aligned} \widehat{V_{01}} &=m_{a} V_{S}=0.8 \times 325=260 \mathrm{~V} \\ \widehat{V_{01}} &=\frac{4 V_{S}}{\pi} \sin d=260 \\ \frac{4(325)}{\pi} \sin d &=260 \\ \sin d &=\frac{260 \times \pi}{4 \times 325}=0.628 \\ d &=38.9 \\ \therefore \qquad \qquad \theta &=\frac{\pi}{2}-d=90^{\circ}-38.9=51.1 \end{aligned}

Q8GATE 2019NAT

The output voltage of a single-phase full bridge voltage source inverter is controlled by unipolar PWM with one pulse per half cycle. For the fundamental rms component of output voltageto be 75% of DC voltage, the required pulse width in degrees (round off up to one decimal place) is _________

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📖 Explanation

\begin{aligned} V_{01,rms}&=\frac{2\sqrt{2}}{\pi}\\ 0.9 \sin d \; V_s&=0.75V_s\\ d&=56.44^{\circ}\\ 2d&=112.88^{\circ}\\ \text{Pulse width }&=112.88^{\circ} \end{aligned}

Q9GATE 2017MCQ

A three-phase voltage source inverter with ideal devices operating in 180 $^{\circ}$ conduction mode is feeding a balanced star-connected resistive load. The DC voltage input is $V_{dc}$ . The peak of the fundamental component of the phase voltage is

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📖 Explanation

$3-\phi \;\text{ VSI }180^{\circ} \text{ mode:}$

\begin{aligned} V_{Rn}&=\frac{6 V_{dc}/3}{n \pi} \sin n\omega t\\ &= \frac{2 \times V_{dc}}{n \pi} \sin n\omega t \\ V_{Rn} &= \frac{2 V_{dc}}{ \pi} \sin \omega t \end{aligned}

Explanation Figure 1

Q10GATE 2017MCQ

The figure below shows a half-bridge voltage source inverter supplying an RL-load with $R=40\Omega$ and $L=(\frac{0.3}{\pi })H$ . The desired fundamental frequency of the load voltage is 50 Hz. The switch control signals of the converter are generated using sinusoidal pulse width modulation with modulation index. M = 0.6. At 50 Hz, the RL-load draws an active power of 1.44 kW. The value of DC source voltage $V_{DC}$ in volts is

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📖 Explanation

\begin{aligned} V_{01 peak}&=M\cdot \frac{V_s}{2} \;\;\;M\rightarrow \text{Modulation index}\\ V_{01peak}&=\frac{M\cdot \frac{V_s}{2}}{\sqrt{2}}=M\cdot \frac{V_s}{2\sqrt{2}}\\ V_{01peak}&=\frac{0.6V_s}{2\sqrt{2}} \end{aligned}

\begin{aligned} V_{01}&=\frac{0.3}{\sqrt{2}}V_s\\ |Z_1|&=\sqrt{R^{2+}(\omega L)^2}\\ &=\sqrt{40^{2+}(2 \pi f L)^2}\\ &=\sqrt{40^{2+}\left (2 \pi \times 50 \times \frac{0.3}{\pi} \right )^2}\\ &=\sqrt{40^{2+}30^2}=50\\ \phi _1&=\tan ^{-1}\frac{\omega L}{R}=\tan ^{-1}\frac{30}{40}=36.896^{\circ}\\ \text{Active power} &=V_{01}I_{01} \cos \phi \\ &=V_{01}\frac{V_{01}}{|Z_1|} \cos \phi \\ 1.44 \times 10^3&=\left (\frac{0.3}{\sqrt{2}}V_s \right )^2\cdot \frac{1}{50}(\cos 36.896^{\circ} )\\ \therefore \;\;V_s^2&=\frac{1.44 \times 10^3 \times 100}{0.3^2 (0.8)}\\ &=20 \times 10^5\\ V_s&=\sqrt{2} \cdot 10^3=1000\sqrt{2}\\ V_{DC}&=\frac{V_s}{2}=500\sqrt{2}V \end{aligned}

Explanation Figure 2

Q11GATE 2017NAT

A 3-phase voltage source inverter is supplied from a 600V DC source as shown in the figure below. For a star connected resistive load of 20 $\Omega$ per phase, the load power for 120 $^{\circ}$ device conduction, in kW is __________.

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📖 Explanation

\begin{aligned} V_{rms}&=\sqrt{\frac{1}{\pi}(300)^2\frac{2 \pi}{3}}\\ &=300\sqrt{\frac{2}{3}}=244.94V\\ 3\frac{V^2}{R}&=\frac{3 \times (244.94)^2}{20}=9kW \end{aligned}

Explanation Figure 2

Q12GATE 2017NAT

In the converter circuit shown below, the switches are controlled such that the load voltage $v_o(t)$ is a 400 Hz square wave. The RMS value of the fundamental component of $v_o(t)$ in volts is _______.

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📖 Explanation

$V_{01}=\frac{4V_s}{\sqrt{2} \pi}=\frac{4 \times 220}{\sqrt{2} \pi}=198.06V$

Explanation Figure 2

Q13GATE 2016NAT

The switches T1 and T2 in Figure (a) are switched in a complementary fashion with sinusoidal pulse width modulation technique. The modulating voltage $v_{m}(t) = 0.8 \sin (200 \pi t) V$ and the triangular carrier voltage ( $v_{c}$ ) are as shown in Figure (b). The carrier frequency is 5 kHz. The peak value of the 100 Hz component of the load current ( $i_{L}$ ), in ampere, is ________ .

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📖 Explanation

\begin{aligned} m_a&=0.8\\ (V_{01})_{peak}&=m_a\frac{V_d}{2}\;\;[m_a\leq 1]\\ &=0.8 \times 250=200V\\ (I_{01})_{peak}&=\frac{(V_{01})_{peak}}{Z_1}\\ &=\frac{200}{\sqrt{R^2}+(\omega L)^2}\\ &=\frac{200}{\sqrt{12^{2+}16^2}}=10A \end{aligned}

Q14GATE 2016NAT

A three-phase Voltage Source Inverter (VSI) as shown in the figure is feeding a delta connected resistive load of 30 $\Omega$ /phase. If it is fed from a 600 V battery, with 180 $^{\circ}$ conduction of solid-state devices, the power consumed by the load, in kW, is __________.

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📖 Explanation

\begin{aligned} V_L&=V_{ph}=\sqrt{\frac{2}{3}}V_s\\ V_{ph}&=\sqrt{\frac{2}{3}} \times 600\\ P&=3\frac{V_{ph}^2}{R}=\frac{3 \times \frac{2}{3} \times 600^2}{30}\\ &=24kW \end{aligned}

Q15GATE 2016MCQ

A single-phase full-bridge voltage source inverter (VSI) is fed from a 300 V battery. A pulse of 120 $^{\circ}$ duration is used to trigger the appropriate devices in each half-cycle. The rms value of the fundamental component of the output voltage, in volts, is

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📖 Explanation

\begin{aligned} V_{01(rms)}&=\frac{2\sqrt{2}}{\pi}V_s \cdot \sin d\\ \text{Pulse width} & \text{ where } 2d=120^{\circ}\\ d=60^{\circ}\\ V_{01(rms)}&=\frac{2\sqrt{2}}{\pi}V_s \sin 60^{\circ}\\ &=\frac{2\sqrt{2}}{\pi}V_s \frac{\sqrt{3}}{2}=234V \end{aligned}

Q16GATE 2015NAT

The single-phase full-bridge voltage source inverter (VSI), shown in figure, has an output frequency of 50 Hz. It uses unipolar pulse width modulation with switching frequency of 50 kHz and modulation index of 0.7. For $V_{\in}$ =100 V DC, L=9.55 mH, C=63.66 $\mu$ F, and R=5 $\Omega$ , the amplitude of the fundamental component in the output voltage $V_o$ (in Volt) under steady-state is ________.

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📖 Explanation

The rms value of the fundamental component of output voltage,

\begin{aligned} V_{R1}&=\frac{V_s}{\sqrt{2}}\times m_i \\ &= \frac{100}{\sqrt{2}} \times 0.7=49.49V \end{aligned}

Amplitude of the value is peak value (i.e.) $V_{R_{peak}}=\sqrt{2} \times 49.49=70V$ $V_{R_{peak}}$ is the amplitude of fundamental voltage at the VSI. $V_{0_{(peak)}}$ is the amplitude of fundamental voltage at the resistor.

\begin{aligned} Z&=\frac{R(-jX_C)}{R-X_Cj} \\ &= \frac{5(-j50)}{5-j50}\\ &= 4.97\angle -5.71^{\circ}\Omega \\ \text{where, } X_C&= \frac{1}{\omega C}\\ &=\frac{1}{2 \pi \times 50 \times 63.66 \times 10^{-6}} \\ &= 50\Omega \\ X_L &=\omega L \\ &=2 \pi \times 50 \times 9.55 \times 10^{-3} \\ &= 3 \Omega\\ V_{0(peak)}&=V_{R_{peak}} \times \frac{4.97\angle -5.71^{\circ}}{(4.97\angle -5.71^{\circ})+j3}\\ &=62.75\angle -32.5^{\circ}\;V \end{aligned}

Therefore, the amplitude of the fundamental component in the output voltage, $V_0=62.75V$

Explanation Figure 2

Q17GATE 2014MCQ

A single-phase voltage source inverter shown in figure is feeding power to a load. The triggering pulses of the devices are also shown in the figure. If the load current is sinusoidal and is zero at $0,\pi ,2\pi..,$ the node voltage $V_{AO}$ has the waveform

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📖 Explanation

$\theta \leq \omega t \leq \pi-\theta \;\;(S_{11}S_4\rightarrow ON)$

\begin{aligned} V_A &=V_{DC}, \; V_0=\frac{V_{DC}}{2} \\ V_{AO}&=\frac{V_{DC}}{2}\;\; (\pi-\theta \lt \omega t \lt z) \end{aligned}

\begin{aligned} V_A &=0\\ V_{AO}&=V\cdot 0-\frac{V_{DC}}{2}\\ &=-\frac{V_{DC}}{2} \end{aligned}

$\pi +\theta \leq \omega t\leq 2 \pi$ $V_A=0,\;\; V_0=\frac{V_d}{2}$

Explanation Figure 3

Explanation Figure 4

Explanation Figure 5

Q18GATE 2014NAT

The figure shows one period of the output voltage of an inverter. $\alpha$ should be chosen such that $60^{\circ} \lt \alpha \lt 90^{\circ}$ . If rms value of the fundamental component is 50V, then $\alpha$ in degree is _____.

🚀 Solve in Practice Mode

📖 Explanation

The output voltage $V_0$ can be reprresnted by fourier series as under:

\begin{aligned} V_0&=\sum_{n=1,3,5,...}^{\infty }a_n \cos n\omega t+b_n \sin n\omega t\\ \text{where, } a_n&=\frac{2}{T}\int_{0}^{T}f(t)\cos n\omega t \;d(\omega t) \\ b_n&=\frac{2}{T}\int_{0}^{T}f(t)\sin n\omega t \;d(\omega t) \\ a_n&=\frac{1}{\pi} \$[\int_{0}^{\alpha }100 \cos \omega t \;d(\omega t) +\int_{\alpha }^{180^{\circ}-\alpha }(-100) \cos \omega t \;d(\omega t) \\ &+ \int_{180^{\circ}-\alpha}^{180^{\circ} }100 \cos \omega t \;d(\omega t) +\int_{180^{\circ} }^{180^{\circ}+\alpha }(-100) \cos \omega t \;d(\omega t)\\ &+ \int_{180^{\circ}+\alpha}^{360^{\circ} -\alpha }100 \cos \omega t \;d(\omega t) +\int_{360^{\circ}-\alpha }^{360^{\circ} }(-100) \cos \omega t \;d(\omega t)]\$\\ a_n &=\frac{100}{\pi}\$[\sin \alpha - \sin (180^{\circ} -\alpha ) + \sin \alpha \\ &+\sin (180^{\circ}) -\sin (180^{\circ} -\alpha )\\ &- \sin (180^{\circ} +\alpha ) + \sin 180^{\circ}\\ &+ \sin (360^{\circ} -\alpha )- \sin (180^{\circ} +\alpha ) \\ &- \sin (360^{\circ} ) + \sin (360^{\circ} -\alpha )]\$\\ a_n &=0 \\ b_n&=\frac{1}{\pi} \$[\int_{0}^{\alpha }100 \sin \omega t \;d(\omega t) +\int_{\alpha }^{180^{\circ}-\alpha }(-100) \sin \omega t \;d(\omega t) \\ &+ \int_{180^{\circ}-\alpha}^{180^{\circ} }100 \sin \omega t \;d(\omega t) +\int_{180^{\circ} }^{180^{\circ}+\alpha }(-100) \sin \omega t \;d(\omega t)\\ &+ \int_{180^{\circ}+\alpha}^{360^{\circ} -\alpha }100 \sin \omega t \;d(\omega t) +\int_{360^{\circ}-\alpha }^{360^{\circ} }(-100) \sin \omega t \;d(\omega t)]\$\\ b_n &=\frac{100}{\pi}\$[-\cos \alpha +1+ \cos (180^{\circ} -\alpha ) - \cos \alpha \\ &-\cos (180^{\circ}) +\cos (180^{\circ} -\alpha )\\ &+ \cos (180^{\circ} +\alpha ) - \cos 180^{\circ}\\ &- \cos (360^{\circ} -\alpha )+ \cos (180^{\circ} +\alpha ) \\ &+ \cos (360^{\circ} ) - \cos (360^{\circ} -\alpha )]\$\\ b_n&=\frac{100}{\pi}[4-8 \cos \alpha ] =50\sqrt{2}\\ \cos \alpha &=0.22231=77.15^{\circ} \end{aligned}

Q19GATE 2013MCQ

The Voltage Source Inverter (VSI) shown in the figure below is switched to provide a 50 Hz, square-wave ac output voltage ( $v_o$ ) across an R-L load. Reference polarity of $v_o$ and reference direction of the output current $i_o$ are indicated in the figure. It is given that R = 3 $\Omega$ , L = 55.9 mH. Appropriate transition i.e., Zero Voltage Switching (ZVS)/Zero Current Switching (ZCS) of the IGBTs during turn-on/turn-off is

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Q20GATE 2012MCQ

In the 3-phase inverter circuit shown, the load is balanced and the gating scheme is 180 $^{\circ}$ -conduction mode. All the switching devices are ideal. If the dc bus voltage $V_d$ =300 V, the power consumed by 3-phase load is

🚀 Solve in Practice Mode

📖 Explanation

Power consumed by each resistor $P=\frac{V_p^2}{R}=\frac{(141.4)^2}{20}=1000W$ Total power consumed =3P=3kW

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