GATE EE · Electrical Machines

Dc Machines Practice Questions

Generate GATE-level questions on DC Machines. Focus on: 1. Working principle of Generator and Motor, EMF and Torque equations. 2. Types of DC machines and their characteristics. 3. Speed control and Braking of DC motors.

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

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Q1GATE 2025MCQ

A DC series motor with negligible series resistance is running at a certain speed driving a load, where the load torque varies as cube of the speed. The motor is fed from a 400 V DC source and draws 40 A armature current. Assume linear magnetic circuit. The external resistance, in $\Omega$ , that must be connected in series with the armature to reduce the speed of the motor by half, is closest to

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📖 Explanation

$T_{e m} \propto I_{a}{ }^{2}$ $\mathrm{T}_{\mathrm{L}} \propto \mathrm{N}^{3}$ $\Rightarrow \mathrm{I}_{\mathrm{a}}{ }^{2} \propto \mathrm{~N}^{3}$ $\left(\frac{\mathrm{I}_{\mathrm{a} 2}}{\mathrm{I}_{\mathrm{a} 1}}\right)^{2}=\left(\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}\right)^{3}$ $\Rightarrow \mathrm{I}_{\mathrm{a} 2}^{2}=40^{2} \times \frac{1}{8}$ $\Rightarrow \mathrm{I}_{\mathrm{a} 2}=14.142 \mathrm{~A}$ $\mathrm{E}_{\mathrm{b} 2}=400-14.14\left(\mathrm{R}_{\mathrm{ex}}\right)$ $\therefore \frac{N_{2}}{N_{1}}=\frac{E_{b 2}}{E_{b 1}} \times \frac{\phi_{1}}{\phi_{2}}$ $\Rightarrow \frac{1}{2}=\frac{400-14.14\left(R_{\text {ex }}\right)}{400} \times \frac{40}{14.142}$ $\Rightarrow \mathrm{R}_{\mathrm{ex}}=23.28 \Omega$

Q9GATE 2018NAT

A separately excited dc motor has an armature resistance $R_{a}= 0.05\Omega$ . The field excitation is kept constant. At an armature voltage of 100 V, the motor produces a torque of 500 Nm at zero speed. Neglecting all mechanical losses, the no-load speed of the motor (in radian/s) for an armature voltage of 150 V is _____ (up to 2 decimal places).

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📖 Explanation

Given separately initiated DC motor, Field excitaion is constant, Producing a torque of 500 N-m

\begin{aligned} &\text{At zero speed,}\\ \because \; N&= 0,\; E_b=0\\ E_b &= V-I_aR_a\\ 0&= 100-I_a(0.05)\\ \therefore \; I_{a1} &=\frac{100}{0.05} =2000A\\ T &= \frac{60}{2 \pi N}E_bI_a\\ &= \frac{60}{2 \pi N}\frac{\phi ZNP}{60A}\cdot I_a\\ \text{Let,}\;\; T_1&=\frac{1}{2 \pi}\frac{ZP}{A}\phi I_{a1}\;\text{N-m} \\ \text{Let,}\;\; T_1 &=kI_{a1} \\ \text{where,}\;\; k &= \frac{1}{2 \pi}\frac{ZP \phi}{A}\\ T_1 &= 500 \text{N-m}, \\ I_{a1}&=2000A\\ k&=\frac{500}{2000}=0.25 \end{aligned}

When motor runs on no-load given all mechanical losses neglected. No-load current is negligible and the voltage drop at no-load can be negligible.

\begin{aligned} \therefore \;\;E_b &\simeq V=150V \\ E_b&= \frac{\phi ZNP}{60A}\\ \text{Already solved,}\\ k&=\frac{1}{2\pi} \frac{ZP \phi }{A}\\ \therefore \;\;\frac{ZP \phi }{A} &=2 \pi (k) \\ \therefore \;\; E_b &= 2 \pi (k) \cdot \frac{N}{60}\\ \therefore \;\; E_b&= k \times \omega \\ (\omega : & \text{No-load speed \in rad/sec})\\ \therefore \;\;\omega &= \frac{E_b}{k}\\ \therefore \;\;\omega &= \frac{150}{0.25}\\ \omega &= 600 \text{rad/sec} \end{aligned}

Q20GATE 2015NAT

A separately excited DC motor runs at 1000 rpm on no load when its armature terminals are connected to a 200V DC source and the rated voltage is applied to the field winding. The armature resistance of this motor is 1 $\Omega$ . The no-load armature current is negligible. With the motor developing its full load torque, the armature voltage is set so that the rotor speed is 500 rpm. When the load torque is reduced to 50% of the full load value under the same armature voltage conditions, the speed rises to 520 rpm. Neglecting the rotational losses, the full load armature current (in Ampere) is _______.

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📖 Explanation

$V_t=200V, N=1000 \text{rpm}, I_{a0}=0A\;...(1)$ $V_t=V(say), N=500\text{rpm}, T_d=T_{fL}\;...(2)$ $V_t=V(said),N=520\text{rpm},$ $T'_L=\frac{T_L}{2}=\frac{T_{FL}}{2}\;\;...(3)$ $P_{lossws}=0$ $R_a=1\Omega$ From condition (1), $E_b=V_t-I_aR_a$ $E_{b_1}=200V, N=1000\text{rpm}$ Since from (2),

\begin{aligned} N_2 &=500\text{rpm} \\ E_{b_2} &=\frac{500 \times 200}{1000}=100V \\ E_{b_2} &= V-I_{fL}R_a\\ 100 &= V-I_{fL}R_a\\ 100 &= V-I_{fL}\;...(4)\\ & \text{Again (3)},\\ N_3 &= 520 \text{rpm}\\ E_{b_3}&= \frac{520 \times 200}{1000}=104V\\ & \text{as we know}\\ 104&=V-\frac{I_{fL}}{2}\times R \\ &= V-\frac{I_{fL}}{2}\;...(5)\\ &\text{From equation (4) and (5),}\\ 100 + I_{fL}&= 104+\frac{I_{fL}}{2}\\ \text{on solving,} I_{fL}&=8A \end{aligned}