GATE 2023 IN Q.51: solution and answer

Official answer: 1.77 to 1.78Source: Official answer key, GATE 2023 IN (IIT Kanpur)

1. Treat the winding as 72 equal turns in series across the 3.6 V supply, with the ground lead at the start of turn 1 and the 3.6 V lead at the end of turn 72. Each turn then drops $$\Delta V=\frac{3.6\text{V}}{72}=0.05\text{V}.$$

2. When the wiper touches a single turn $n$, there are $n$ turns between it and ground, so $V_n=n\Delta V$. This scale reads $V_{72}=3.6\text{V}$ on the last turn. So $V_{35}=35\times 0.05=1.75\text{V}$ and $V_{36}=36\times 0.05=1.80\text{V}$.

3. Make-before-break means the wiper is two turns wide. On turns 35 and 36 together it touches both contact points at once, so it shorts the one turn of wire between them (turn 36). With no load on the wiper, that turn carries no current and drops out of the divider.

4. The divider now has $71$ turns in circuit: $35$ turns (1 to 35) between the wiper and ground, and $36$ turns (37 to 72) between the wiper and the 3.6 V end. $$V_o=3.6\times \frac{35}{71}=1.7746\text{V}.$$

5. Check: $1.7746\text{V}$ sits almost exactly halfway between the single-turn readings $1.75\text{V}$ and $1.80\text{V}$ (their mean is $1.775\text{V}$). That is the point of a two-turn wiper: it adds a half-turn step between neighbouring turns and so doubles the resolution.

6. $V_o\approx 1.77\text{V}$ (rounded to two decimal places).