Question

The no-load steady-state output voltage of a DC shunt generator is 200 V when it is driven in the clockwise direction at its rated speed. If the same machine is driven at the rated speed but in the opposite direction, the steady-state output voltage will be ______ V (rounded off to the nearest integer).

Hint:

A DC shunt generator will not self-excite if you reverse rotation without swapping field (or brush) connections. To generate in reverse, reverse the shunt-field leads.

Answer 1

Correct Answer: 0

Step 1: Self–excitation condition. For a shunt generator to build up, the field connection must be such that the current produced by the {residual} emf reinforces the residual flux. The generated emf is $E = k\,\Phi\,\omega$ (sign depends on rotation).
Step 2: Reverse the direction of rotation. With the same field/brush connections kept unchanged, reversing $\omega$ reverses the polarity of the induced emf. The small residual emf then drives field current in the {opposite} direction, which {opposes} the residual flux, so the cumulative buildup fails and the machine {demagnetizes} instead of exciting.
Step 3: Steady-state outcome. Without build-up, only negligible residual voltage remains; in the ideal setting asked, the terminal no-load voltage tends to 0 V.
Final Answer: 0 V