GATE 2018 EE Q.31: solution and answer

Official answer: BSource: Official answer key, GATE 2018 EE (IIT Guwahati)

1. Set up the loop equations for $t>0$. With S closed, the current $i(t)$ flows clockwise: out of the source's + terminal, through $L$, and into the + terminal of $C$. That gives the passive sign convention at the capacitor, so $i=C\frac{dv}{dt}$. KVL around the loop gives $L\frac{di}{dt}+v=5$. With $L=1\text{H}$ and $C=1\text{F}$, substitute $i=\frac{dv}{dt}$: $$\frac{d^{2}v}{d{t}^2}+v=5,{\omega }_0=\frac{1}{\sqrt{LC}}=1\text{rad/s}$$

2. General solution (particular plus natural): $v(t)=5+A\cos t+B\sin t$.

3. Apply the initial conditions. $v(0)=0$ gives $5+A=0$, so $A=-5$. $i(0)=C{v}^{\prime }(0)=B=0$. Hence $$v(t)=5(1-\cos t)\text{V}$$

4. Current: $i(t)=C\frac{dv}{dt}=5\sin t\text{A}$. Check: at $t=0^{+}$, $\frac{di}{dt}=\frac{5-v(0)}{L}=5\text{A/s}$, which agrees with $\frac{d}{dt}(5\sin t){|}_0=5$.

5. Eliminate $t$. From the two results, $\cos t=\frac{5-v}{5}$ and $\sin t=\frac{i}{5}$. Then ${\sin }^{2}t+{\cos }^{2}t=1$ gives $${\left(\frac{v-5}{5}\right)}^2+{\left(\frac{i}{5}\right)}^2=1\Rightarrow (v-5{)}^2+i^2=25$$

6. Cross-check with energy (the circuit is lossless). Multiply $L\frac{di}{dt}=5-v$ by $i=C\frac{dv}{dt}$ to get $Li\frac{di}{dt}=C(5-v)\frac{dv}{dt}$. Integrate from $t=0$, where $i=0$ and $v=0$: $$\frac{1}{2}L{i}^2=C\left(5v-\frac{1}{2}{v}^2\right)\Rightarrow i^2=10v-v^2\Rightarrow (v-5{)}^2+i^2=25$$ This is the same circle.

7. Read off the locus. It is a circle of radius $5$ centred at $(v,i)=(5,0)$. It passes through the origin at $t=0$ and touches the $i(t)$ axis there. Over each cycle, $v$ swings from $0$ to $10\text{V}$ and $i$ from $-5$ to $+5\text{A}$. This is option (B).