GATE EE · Power Electronics

Electrical Drives – GATE EE Power ElectronicsPractice Questions & PYQs

Generate GATE-level questions on Electrical Drives. Focus on: 1. Speed control of DC motors using Controlled Rectifiers and Choppers. 2. Speed control of Induction motors: V/f control, Rotor resistance control, and Slip power recovery. 3. Four-quadrant operation and regenerative braking.

11 questions · 11 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2024NAT

A single-phase full-controlled thyristor converter bridge is used for regenerative braking of a separately excited DC motor with the following specifications:

\begin{array}{|l|l|} \hline \text{Rated armature voltage} & 210 \mathrm{~V} \\ \hline \text{Rated armature current} & 10 \mathrm{~A} \\ \hline \text{Rated speed} & 1200 \mathrm{rpm} \\ \hline \text{Armature resistance} & 1 \Omega \\ \hline \text{Input to the converter bridge} & 240 \mathrm{~V} at 50 \mathrm{~Hz} \\ \hline \begin{array}{l} \text{The armature of the DC motor is fed from the} \\ \text{full-controlled bridge and the field current is kept} \\ \text{constant.} \\ \end{array} & \\ \hline \end{array}

Assume that the motor is running at $600 \mathrm{rpm}$ and the armature terminals of the motor are suitably reversed for regenerative braking. If the armature current of the motor is to be maintained at the rated value, the triggering angle of the converter bridge in degrees should be _____ (rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

Rating of DC machine 210 V, 10 A, 1200 rpm,

\begin{aligned} & R_{a}=1 \Omega \\ & I_{F} \rightarrow \text { constant } \\ & \therefore \quad \phi \rightarrow \text { constant } \\ & \therefore \quad E_{b}=K_{m} W=K_{m} \cdot \frac{2 \pi \cdot N}{60} \\ & V_{0}=E_{b}+I_{0} R_{a} \\ & 210=K_{m} \cdot \frac{2 \pi N}{60}+I_{0} R_{a} \\ & 210=K_{m} \cdot \frac{2 \pi \cdot 1200}{60}+(10 \times 1) \\ & K_{m}=1.59(\mathrm{~V}-\mathrm{s}) / \mathrm{rad} \end{aligned}

Armature terminals reversed $\therefore \quad E_{b} \rightarrow$ Negative Inv for regenerative braking

\begin{aligned} V_{0} & =-E_{b}+I_{0} R_{a} \\ \frac{2 V_{m}}{\pi} \cos \alpha & =-E_{b}+I_{0} R_{a} \end{aligned}

\begin{aligned} \frac{2.240 \sqrt{2}}{\pi} \cos \alpha & =-K_{m} \cdot \frac{2 \pi N}{60}+I_{0} R_{a} \\ \frac{2 \cdot 240 \sqrt{2}}{\pi} \cdot \cos \alpha & =-1.59\left(\frac{2 \pi \cdot 600}{60}\right)+10 \\ \alpha & =114.58^{\circ} \end{aligned}

Q2GATE 2015NAT

A shunt-connected DC motor operates at its rated terminal voltage. Its no-load speed is 200 radian/second. At its rated torque of 500 Nm, its speed is 180 radian/second. The motor is used to directly drive a load whose load torque $T _{L}$ depends on its rotational speed $\omega _{r}$ (in radian/second), such that $T _{L}=2.78 \times \omega _{r}$ . Neglecting rotational losses, the steady-state speed (in radian/second) of the motor, when it drives this load, is _______.

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📖 Explanation

$T_L=2.78 \times \omega$ After neglecting rotational losses, $(T_{sh}=T_a)=T_L$ At rated load,

\begin{aligned} T_a&=500Nm=2.78 \times \omega \\ \omega _r&=179.85 \\ \omega _r&\simeq 180 \text{ rad/sec} \end{aligned}

Q3GATE 2013MCQ

The separately excited dc motor in the figure below has a rated armature current of 20 A and a rated armature voltage of 150 V. An ideal chopper switching at 5 kHz is used to control the armature voltage. If $L_a=0.1mH, R_a=1\Omega$ , neglecting armature reaction, the duty ratio of the chopper to obtain 50% of the rated torque at the rated speed and the rated field current is

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📖 Explanation

\begin{aligned} V_a &=E_b+I_aR_a \\ 150&=E_b+20 \times 1 \\ E_b&= 130V\\ \text{For half} &\text{ the rated torque,} \\ I_a&=10A \\ V_a&=130+10 \times 1=140 \\ D&=\frac{140}{200}=0.7 \end{aligned}

Q4GATE 2009MCQ

A 3-phase squirrel cage induction motor supplied from a balanced 3-phase source drives a mechanical load. The torque-speed characteristics of the motor(solid curve) and of the load(dotted curve) are shown. Of the two equilibrium points A and B, which of the following options correctly describes the stability of A and B ?

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📖 Explanation

Let torque developed by the motor $=T_m$ Load torque $=T_L$ Accelerating torque, $T_a=T_m-T_L$ At point A, $T_m=T_L$ $\Rightarrow \;\;T_a=T_m-T_L=0$ So, no accelerating , hence the motor rums at speed $N_A$ . When a smal disturbance decreases speed to $N_c$ At point C, $T_m \gt T_L$ $\Rightarrow \;\;T_a \gt 0$ So, motor will accelerate and comes to point A. When the disturbance increases speed to $N_D$ . At point D, $T_m \lt T_L$ $\Rightarrow \;\;T_a \lt 0$ Motor will decelerate and comes at point A So, motor is stable at point A. When the motor is operating at point B., $T_m = T_L$ $\Rightarrow \;\;T_a= 0$ If a small disturbance decreases speed to $N_E$ . $T_m \lt T_L$ $\Rightarrow \;\;T_a \lt 0$ Motor will decelerate and motor speed will keep on decreasing. If the disturbance increases speed to $N_F$ . At point F, $T_m \gt T_L$ $\Rightarrow \;\;T_a \gt 0$ Motor will accelerate and motor speed will keep on increasing. So, motor is unstable at point B.

Q5GATE 2008MCQ

A single phase fully controlled converter bridge is used for electrical braking of a separately excited dc motor. The dc motor load is represented by an equivalent circuit as shown in the figure. Assume that the load inductance is sufficient to ensure continuous and ripple free load current. The firing angle of the bridge for a load current of $I_0=10 A$ will be

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📖 Explanation

Average output voltage of the converter, $V_0=\frac{2V_m}{\pi}\cos \alpha$ Load current $=I_0=10A$ Back emf $=E_b=150V$ Armature resistance $=R_a=2\Omega$ Applying KVL,

\begin{aligned} &V_0-2I_0+150=0 \\ \Rightarrow \;& V_0=-150+2 \times 10\\ &=-130V\\ \Rightarrow \; & \frac{2V_m}{\pi} \cos \alpha =-130\\ \Rightarrow \; &\frac{2 \times \sqrt{2} \times 230}{\pi}\cos \alpha=-130 \\ \Rightarrow \;&\alpha =129^{\circ} \end{aligned}

Q6GATE 2007MCQ

A three-phase, 440 V, 50 Hz ac mains fed thyristor bridge is feeding a 440 V dc, 15 kW, 1500 rpm separately excited dc motor with a ripple free continuos current in the dc link under all operating conditions, Neglecting the losses, the power factor of the ac mains at half the rated speed is

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📖 Explanation

For a separately excited dc motor back emf $=E_a=V_0-I_aR_a$ Since, losses are neglected $R_a$ can be neglected So,

\begin{aligned} E_a&=V_0 \\ V_0&=E_a =k_a\phi N \;\;...(i)\\ V_0&\propto N \end{aligned}

At rated voltage $V_0=440V$ and N=1500rpm, so, at half the rated speed $\left ( \frac{N}{2} =750 \;rpm\right )$ output voltage of the bridge ( $V_0$ ) is 220V. if $I_a$ is the average value of armature current rms value of supply current will be $I_s=I_a\sqrt{\frac{2}{3}}$ Power delivered to the motor $P_0=V_0I_a$ Input VA to the thyristor bridge $S_{\in}=\sqrt{3}V_sI_s$ Input power factor $=\frac{P_0}{S_{\in}}=\frac{V_0I_a}{\sqrt{3}V_sI_s}=\frac{220 \times I_a}{\sqrt{3} \times 440 \times I_a\sqrt{\frac{2}{3}}}=0.345$

Q7GATE 2006MCQ

The speed of a 3-phase, 440 V, 50 Hz induction motor is to be controlled over a wide range from zero speed to 1.5 time the rated speed using a 3-phase voltage source inverter. It is desired to keep the flux in the machine constant in the constant torque region by controlling the terminal voltage as the frequency changes. The inverter output voltage vs frequency characteristic should be

🚀 Solve in Practice Mode

📖 Explanation

For a $3-\phi$ induction motor, stator voltage per phase is given by

\begin{aligned} V_1 &=\sqrt{2}\pi f_1 N_{pn_1}\phi K_{\omega 1} \\ \phi &\propto \frac{V_1}{f_1} \end{aligned}

To keep the $\phi$ constant, constant V/f control is applied and voltage ( $V_1$ ) varies linearly with frequency ( $f_1$ ). For frequency above 50 Hz, such control is not possible because voltage can not be increased above rated voltage. Therefore, for frequency above 50 Hz, voltage is kept constant, During this control induction motor is said to be working in field weakling mode.

Q8GATE 2006MCQ

A solar cell of 350 V is feeding power to an ac supply of 440 V, 50 Hz through a 3-phase fully controlled bridge converter. A large inductance is connected in the dc circuit to maintain the dc current at 20 A. If the solar cell resistance is 0.5 $\Omega$ , then each thyristor will be reverse biased for a period of

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📖 Explanation

Solar cell emf = 350V DC current $I_{dc}=20A$ Solar cell resistance

\begin{aligned} R_{cell}&=0.5\Omega \\ V_0&=\text{Voltage across inverter} \\ &=-(E-I_{dc}R_{cell}) \\ &=-(350-20 \times 0.5)=-340V \end{aligned}

The bridge acts as inverter, Output voltage of $3-\phi$ fully controlled bridge

\begin{aligned} V_0&=\frac{3V_{ml} \cos \alpha }{\pi}\\ -340&=\frac{3V_{ml} \cos \alpha }{\pi}\\ -340&=\frac{3 \times 440\sqrt{2} \cos \alpha }{\pi}\\ \Rightarrow \; \alpha &=125^{\circ} \end{aligned}

Therefore, each thyristor will be reverse biased for a period of $55^{\circ}$ .

Q9GATE 2005MCQ

An electric motor, developing a starting torque of 15 Nm, starts with a load torque of 7 Nm on its shaft. If the acceleration at start is 2 rad/ $sec^{2}$ , the moment of inertia of the system must be (neglecting viscous and coulomb friction)

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📖 Explanation

$T_s=$ Starting torque developed by the motor =15 N-m $T_L=$ Load torque =7 N-m $T_a=$ Accelerating torque $=T_s-T_l=$ 15-7=8 N-m $\alpha =$ Acceleration $= 2 \; rad/sec^2$ $T_a=I\alpha$ $I=$ Moment of inertia $=\frac{T_a}{\alpha }=\frac{8}{2}=4kgm^2$

Q10GATE 2004MCQ

A single-phase half-controlled rectifier is driving a separately excited dc motor. The dc motor has a back emf constant of 0.5 V/rpm. The armature current is 5A without any ripple. The armature resistance is 2 $\Omega$ . The converter is working from a 230V, single-phase ac source with a firing angle of $30^{\circ}$ . Under this operating condition, the speed of the motor will be

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \text{Back emf }&=E_a=k\phi N\\ E_a&=k_bN \end{aligned}

where, $k_b=$ Back-emf constant =0.25 V/rpm Average output voltage of $1-\phi$ half controlled rectifier =V

\begin{aligned} V &=\frac{V_m}{\pi} (1+ \cos \alpha ) \\ &= \frac{230\sqrt{2}}{2 \pi} (1+ \cos 30^{\circ})\\ \Rightarrow \; V &=96.6V \\ E_a&=V-I_aR_a \\ &= 96.6-5 \times 2=86.6V\\ \text{Speed}&=N=\frac{E_a}{k_b}=\frac{86.6}{0.25}=346.4V \end{aligned}

Q11GATE 2004MCQ

A variable speed drive rated for 1500 rpm, 40 Nm is reversing under no load. Figure shows the reversing torque and the speed during the transient. The moment of inertia of the drive is

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📖 Explanation

Speed changes from -1500 rpm to 500 rpm in 0.5 sec So angular acceleration,

\begin{aligned} \alpha &=\frac{500-(-1500)}{0.5}\times \frac{2 \pi}{60}\\ &=418.88 \;rad/sec^2\\ \text{Torque}&=T=20N-m\\ T&=I\alpha \\ &\text{Moment of inertia}\\ I&=\frac{T}{\alpha }=\frac{20}{418.88}=0.048kgm^2 \end{aligned}

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Electrical Drives – GATE EE Power ElectronicsPractice Questions & PYQs

11 questions · 11 PYQs · 0 AI practice · GATE EE 2027

Q1GATE 2024NAT

A single-phase full-controlled thyristor converter bridge is used for regenerative braking of a separately excited DC motor with the following specifications:

\begin{array}{|l|l|} \hline \text{Rated armature voltage} & 210 \mathrm{~V} \\ \hline \text{Rated armature current} & 10 \mathrm{~A} \\ \hline \text{Rated speed} & 1200 \mathrm{rpm} \\ \hline \text{Armature resistance} & 1 \Omega \\ \hline \text{Input to the converter bridge} & 240 \mathrm{~V} at 50 \mathrm{~Hz} \\ \hline \begin{array}{l} \text{The armature of the DC motor is fed from the} \\ \text{full-controlled bridge and the field current is kept} \\ \text{constant.} \\ \end{array} & \\ \hline \end{array}

🚀 Solve in Practice Mode

📖 Explanation

Rating of DC machine 210 V, 10 A, 1200 rpm,

\begin{aligned} & R_{a}=1 \Omega \\ & I_{F} \rightarrow \text { constant } \\ & \therefore \quad \phi \rightarrow \text { constant } \\ & \therefore \quad E_{b}=K_{m} W=K_{m} \cdot \frac{2 \pi \cdot N}{60} \\ & V_{0}=E_{b}+I_{0} R_{a} \\ & 210=K_{m} \cdot \frac{2 \pi N}{60}+I_{0} R_{a} \\ & 210=K_{m} \cdot \frac{2 \pi \cdot 1200}{60}+(10 \times 1) \\ & K_{m}=1.59(\mathrm{~V}-\mathrm{s}) / \mathrm{rad} \end{aligned}

Armature terminals reversed $\therefore \quad E_{b} \rightarrow$ Negative Inv for regenerative braking

\begin{aligned} V_{0} & =-E_{b}+I_{0} R_{a} \\ \frac{2 V_{m}}{\pi} \cos \alpha & =-E_{b}+I_{0} R_{a} \end{aligned}

\begin{aligned} \frac{2.240 \sqrt{2}}{\pi} \cos \alpha & =-K_{m} \cdot \frac{2 \pi N}{60}+I_{0} R_{a} \\ \frac{2 \cdot 240 \sqrt{2}}{\pi} \cdot \cos \alpha & =-1.59\left(\frac{2 \pi \cdot 600}{60}\right)+10 \\ \alpha & =114.58^{\circ} \end{aligned}