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 featuring a parallel plate capacitor, the alternating current (AC) source charges the capacitor. This analysis demonstrates that the conduction current (\( I_c \)) and the displacement current (\( I_d \)) are equivalent at all circuit locations. The derivation proceeds stepwise.

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

Current from the AC source charges the capacitor. The conduction current (\( I_c \)) represents this source current flowing to the capacitor. Due to the time-varying AC voltage, this current also varies sinusoidally.

The conduction current \( I_c \) relates to the AC source voltage \( V(t) \). For a sinusoidal source, the current is determined by Ohm's law or capacitive reactance:

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

where:

• \( I_c(t) \) is the conduction current at time \( t \),
• \( C \) is the capacitor's capacitance, and
• \( \frac{dV(t)}{dt} \) is the rate of change of voltage across the capacitor.

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

Displacement current accounts for the time-varying electric field within the capacitor. As the AC source alters the capacitor's charge, the electric field between its plates changes.

The displacement current \( I_d \) is defined as:

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

where:

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

The electric field \( E(t) \) between the plates is \( E(t) = \frac{V(t)}{d} \), where \( d \) is the plate separation. Thus, the displacement current can be expressed as:

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

Since \( d \) is constant, this simplifies to:

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

3. Relating Conduction Current and Displacement Current:

Comparing the expressions for \( I_c \) and \( I_d \):

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

and

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

The capacitance \( C \) is given by \( C = \epsilon_0 \frac{A}{d} \).

Substituting this into the \( I_c \) equation yields:

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

Comparing this with the expression for \( I_d \), we find:

\[ I_c = I_d \]

4. Conclusion:

Consequently, the conduction current \( I_c \) and the displacement current \( I_d \) are equal at all points in the circuit. The conduction current charges the capacitor, while the displacement current arises from the changing electric field within it. Both are governed by the time rate of change of voltage, ensuring their equality throughout the circuit.