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GATE 2015 EE – Question 31

Electrical Machines · DC machines: separately excited, series and shunt, motoring and generating mode of operation and their characteristics, speed control of dc motors · 1 mark · Multiple choice

A separately excited DC generator has an armature resistance of $0.1\ \Omega$ and negligible armature inductance. At rated field current and rated rotor speed, its open-circuit voltage is 200 V. When this generator is operated at half the rated speed, with half the rated field current, an un-charged $1000\ \mu\text{F}$ capacitor is suddenly connected across the armature terminals. Assume that the speed remains unchanged during the transient. At what time (in microsecond) after the capacitor is connected will the voltage across it reach 25 V?

  1. 62.25
  2. 69.3
  3. 73.25
  4. 77.3

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Correct answer: (B) 69.3

Explanation

The generated voltage is proportional to speed and field, so it is $200 \times \frac{1}{2} \times \frac{1}{2} = 50$ V. The capacitor charges through the armature resistance with $\tau = RC = 0.1 \times 1000 \times 10^{-6} = 100\ \mu$s. So $v_c = 50(1 - e^{-t/\tau})$. Setting $v_c = 25$ gives $t = \tau\ln 2 = 69.3\ \mu$s.