GATE 2026 EC Q.23: solution and answer

Official answer: ASource: Official answer key, GATE 2026 EC (IIT Guwahati)

1. Read the equation as KCL at a capacitor node: $C\frac{dV}{dt}$ is the current into a capacitor $C$, $I(t)$ is the source current and $m(t)$ is the current drawn by the rest of the circuit. The impulse $q\delta (t)$ delivers a charge $q$ (in $\text{C}$) in zero time.

2. Integrate both sides across the impulse, from $t=0^{-}$ to $t=0^{+}$: $$C\left[V(0^{+})-V(0^{-})\right]={\int }_{0^{-}}^{0^{+}}q\delta (t)dt-{\int }_{0^{-}}^{0^{+}}m(t)dt$$

3. By the sifting property, ${\int }_{0^{-}}^{0^{+}}q\delta (t)dt=q$. The rest of the circuit draws a finite current (the standard assumption, implicit in the question: $m(t)$ contains no impulse), and a finite function integrated over an interval of zero width gives $0$.

4. So the voltage jumps by the injected charge divided by $C$: $$V(0^{+})=V_0+\frac{q}{C}$$ where $q/C$ is in $\text{V}$.

5. For $t>0$ we have $\delta (t)=0$, so the circuit obeys $C\frac{dV}{dt}=-m(t)$, starting from $V_0+q/C$.

6. Equivalent picture: remove the impulse source and give the capacitor the initial voltage $V_0+q/C$. With no impulse in this equivalent system, $V$ is continuous at $t=0$, so its value at $t=0^{-}$ equals its value at $t=0^{+}$: $V(0^{-})=V_0+q/C$. This is option (A).

7. Check in the Laplace domain (initial conditions taken at $0^{-}$). The original equation transforms to $$C\left[sV(s)-V_0\right]=q-M(s)$$ Moving $q$ to the left: $$C\left[sV(s)-\left(V_0+\frac{q}{C}\right)\right]=-M(s)$$ which is exactly the transform of $C\frac{dV}{dt}=-m(t)$ with $V(0^{-})=V_0+q/C$. Both systems give the same $V(s)$, so they are equivalent.

8. The other options fail: (B) and (C) add $m(0^{-})$, a current in $\text{A}$, to a voltage, which is dimensionally inconsistent; (C) also has the wrong sign on $q/C$ (a positive charge raises the capacitor voltage). (D) drops $V_0$ and is right only when $V_0=0$.