GATE 2023 IN Q.35: solution and answer

Official answer: 0Source: Official answer key, GATE 2023 IN (IIT Kanpur)

1. A DC shunt generator has no independent excitation source; it relies on a small residual flux in the poles. As the armature rotates (clockwise case, given), this residual flux induces a small EMF, which drives a small current through the shunt field winding.

2. Self-excitation ('voltage build-up') occurs only if the mmf produced by this field current reinforces the residual flux: flux rises, EMF rises, field current rises further — a cumulative process that builds the no-load terminal voltage up to 200 V at rated speed (as given for the clockwise direction).

3. The induced EMF's polarity depends on the direction of rotation, since $e\propto \frac{d\varphi }{dt}$ (equivalently $e\propto v\times B$ for the moving conductors) changes sign when the direction of rotation reverses.

4. The physical connection of the field winding to the armature terminals/brushes is fixed hardware and does not change merely because the shaft is driven in the opposite direction.

5. Therefore, running the machine in the opposite (counter-clockwise) direction reverses the polarity of the small residual-flux-generated EMF, and hence reverses the direction of current flowing through the same, unchanged shunt field winding.

6. This reversed field current now produces an mmf that opposes (demagnetizes) the residual flux instead of reinforcing it, because the sense of reinforcement that existed in the clockwise case is tied to the specific combination of rotation direction and fixed winding connection.

7. With an opposing/demagnetizing mmf, cumulative voltage build-up cannot occur — there is no regenerative loop, and the residual flux is driven toward cancellation rather than being reinforced.

8. Hence the machine fails to self-excite when driven in the reverse direction, and the steady-state no-load output voltage is $0\text{V}$ (rounded off to the nearest integer).