GATE 2024 EC Q.23: solution and answer

Official answer: ASource: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Identify the circuit. The secondary has three terminals: top, a centre tap and bottom. $D_1$ runs from the top end and $D_2$ from the bottom end, and their cathodes are joined. $R_L$ sits between the centre tap ($-$) and the joined cathodes ($+$). This is a centre-tapped full-wave rectifier. The figure is drawn symmetric, so the tap is taken at the midpoint of the secondary.

2. Transformer ratio (ideal, $n:1$). The $n:1$ label covers the transformer as a whole, so it is the ratio of primary turns to total secondary turns. The voltage across the whole secondary is $\frac{V_s(t)}{n}=\frac{10}{n}\sin \omega t\text{V}$, with peak $\frac{10}{n}\text{V}$.

3. Half-winding voltage. Each diode works with only half of the secondary, from one end to the centre tap. Let $e(t)=\frac{1}{2}\cdot \frac{10}{n}\sin \omega t=V_m\sin \omega t$, with $V_m=\frac{5}{n}\text{V}$. Measured from the centre tap, the top end is at $+e(t)$ and the bottom end is at $-e(t)$.

4. Conduction. $V_L$ is the cathode node measured from the centre tap. With ideal diodes and $R_L$ as the only return path, whichever end is higher above the centre tap drives the cathode node. So $V_L(t)=\max (e(t),-e(t))=V_m|\sin \omega t|$. When $e>0$, $D_1$ conducts and $D_2$ is reverse biased by $2e$. When $e<0$, the roles swap. No dots are marked on the windings, but either winding polarity gives the same $V_L(t)$.

5. Average of a full-wave rectified sine. The waveform repeats every $\pi$ in $\omega t$: $$V_{L,\text{avg}}=\frac{1}{\pi }{\int }_0^{\pi }{V}_m\sin \theta d\theta =\frac{2V_m}{\pi }=\frac{2}{\pi }\cdot \frac{5}{n}=\frac{10}{n\pi }\text{V}$$

6. Equate to the given value. $$\frac{10}{n\pi }=\frac{2.5}{\pi }\Rightarrow n=\frac{10}{2.5}=4$$ This is option (A).