GATE 2018 IN Q.20: solution and answer

Official answer: DSource: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Read the signal: $v(t)=6\cos (100\pi t)\text{V}$, so the peak value is $V_m=6\text{V}$ and the angular frequency is $\omega =100\pi \text{rad/s}$.

2. Frequency and period from $\omega =2\pi f$: $f=\frac{100\pi }{2\pi }=50\text{Hz}$, so $T=\frac{1}{f}=\frac{1}{50}\text{s}=20\text{ms}$.

3. Read the screen vertically: the positive peak is 3 divisions above the centre line and the negative peak is 3 divisions below it, so the amplitude spans 3 vertical divisions (6 divisions peak-to-peak).

4. Vertical sensitivity = volts per division: $\frac{V_m}{3\text{div}}=\frac{6\text{V}}{3\text{div}}=2\text{V/div}$. Check with peak-to-peak values: $\frac{12\text{V}}{6\text{div}}=2\text{V/div}$.

5. Read the screen horizontally: the trace goes from the centre line up to the peak, down through the centre line at the middle of the screen, down to the trough and back to the centre line at the right edge. That is exactly one complete cycle across all 10 horizontal divisions.

6. Time base = time per division: $\frac{T}{10\text{div}}=\frac{20\text{ms}}{10\text{div}}=2\text{ms/div}$.

7. The trace starts at a zero crossing, not at a peak, only because of where the sweep is triggered. The trigger point sets the starting phase on screen but does not change the amplitude or the period. So the settings are $2\text{V/div}$ and $2\text{ms/div}$, which is option (D).