Question

A parallel plate capacitor is charged by an ac source. Show that the sum of conduction current (\( I_c \)) and the displacement current (\( I_d \)) has the same value at all points of the circuit.

Hint:

In AC circuits, displacement current is equal to the conduction current, which is why Kirchhoff’s current law is still applicable in circuits with capacitors.

Answer 1

In an AC circuit with a parallel plate capacitor, the capacitor is charged by the alternating current (AC) source. In this scenario, we need to show that the sum of the conduction current (\( I_c \)) and the displacement current (\( I_d \)) is the same at all points of the circuit. Let's break down the process and derive the relation step by step.

1. Conduction Current (\( I_c \)) in the Circuit:

When the capacitor is connected to the AC source, the current from the source charges the capacitor. The conduction current (\( I_c \)) is the current supplied by the AC source that flows through the wires leading to the capacitor. This current fluctuates sinusoidally because the voltage supplied by the AC source is time-varying.

The conduction current \( I_c \) in the circuit is related to the voltage \( V(t) \) supplied by the AC source. For a sinusoidal AC source, the current is given by Ohm's law or the capacitive reactance, where the relationship is:

\[ I_c(t) = C \frac{dV(t)}{dt} \]

where:

• \( I_c(t) \) is the conduction current at time \( t \),
• \( C \) is the capacitance of the capacitor, and
• \( \frac{dV(t)}{dt} \) is the time rate of change of the voltage across the capacitor.

2. Displacement Current (\( I_d \)) in the Capacitor:

The displacement current is a concept introduced to account for the time-varying electric field inside the capacitor. In a parallel plate capacitor, the electric field between the plates changes as the charge on the plates changes with time, due to the time-varying voltage from the AC source.

The displacement current \( I_d \) is defined by the relation:

\[ I_d = \epsilon_0 A \frac{dE}{dt} \]

where:

• \( I_d \) is the displacement current,
• \( \epsilon_0 \) is the permittivity of free space,
• \( A \) is the area of each plate of the capacitor, and
• \( \frac{dE}{dt} \) is the time rate of change of the electric field between the plates.

Since the electric field \( E(t) \) between the plates is related to the voltage \( V(t) \) by the equation \( E(t) = \frac{V(t)}{d} \), where \( d \) is the separation between the plates, the displacement current can also be written as:

\[ I_d = \epsilon_0 A \frac{d}{dt} \left( \frac{V(t)}{d} \right) \]

Since \( d \) is constant (the distance between the plates does not change), we can simplify this to:

\[ I_d = \epsilon_0 A \frac{dV(t)}{dt} \]

3. Relating Conduction Current and Displacement Current:

We can now compare the expressions for the conduction current \( I_c \) and the displacement current \( I_d \). From the previous equations:

\[ I_c = C \frac{dV(t)}{dt} \]

and

\[ I_d = \epsilon_0 A \frac{dV(t)}{dt} \]

Since the capacitance \( C \) of the capacitor is related to the area \( A \) of the plates and the distance \( d \) between them by the equation:

\[ C = \epsilon_0 \frac{A}{d} \]

Substituting this into the equation for \( I_c \), we get:

\[ I_c = \frac{A}{d} \frac{dV(t)}{dt} \]

Now, comparing this expression for \( I_c \) with the expression for \( I_d \), we see that:

\[ I_c = I_d \]

4. Conclusion:

Therefore, the conduction current \( I_c \) and the displacement current \( I_d \) have the same value at all points in the circuit. This is because the conduction current in the circuit is responsible for charging the capacitor, and the displacement current arises due to the changing electric field inside the capacitor. Since the electric field and the voltage are related, the time rate of change of the voltage determines both currents, and thus they must be equal at every point in the circuit.