GATE 2022 EC Q.24: solution and answer

Official answer: CSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

1. No charge crosses the gap between the plates of an ideal capacitor, so the current "through" the capacitor is entirely displacement current: $$i_d=\frac{d}{dt}{\int }_S\mathbf{D}\cdot d\mathbf{S},$$ where $S$ is a surface lying in the gap.

2. Continuity of total current: take the divergence of the Ampère–Maxwell law $\nabla \times \mathbf{H}={\mathbf{J}}_c+\frac{\partial \mathbf{D}}{\partial t}$. Since $\nabla \cdot (\nabla \times \mathbf{H})=0$, we get $$\nabla \cdot \left({\mathbf{J}}_c+\frac{\partial \mathbf{D}}{\partial t}\right)=0.$$ So the total current (conduction plus displacement) is the same through every cross-section of a single unbranched loop.

3. In the series loop the resistor and the wires carry only conduction current, and the capacitor gap carries only displacement current. Series elements carry the same total current at every instant, so $$i_d(t)=i_c(t).$$

4. Check at one capacitor plate. If the plate holds charge $q$, the wire's conduction current is $i_c=\frac{dq}{dt}$. By Gauss's law, the flux of $\mathbf{D}$ out of that plate into the gap is ${\int }_S\mathbf{D}\cdot d\mathbf{S}=q$ (for a parallel-plate capacitor, $D=q/A$). Hence $i_d=\frac{d}{dt}{\int }_S\mathbf{D}\cdot d\mathbf{S}=\frac{dq}{dt}=i_c$, and in circuit terms both equal $C\frac{d{v}_C}{dt}$.

5. The equality holds at every instant, so amplitude and phase are both unchanged: $$i_d=2\sin \left(t+\pi /2\right)\text{A}.$$ This is option (C).