Q137 / 180 · COMEDK 2024 (Morning Shift) · Physics
The capacitance of a parallel plate capacitor is $400 \mathrm{~pF}$. It is connected to an ac source of $100 \mathrm{~V}$ having an angular frequency $100 \mathrm{~rad~s}^{-1}$. If the rms value of the current is $4 \mu \mathrm{A}$, the displacement current is:
- Correct answerA4 $\mu$A
- B$4 \times 10^{-2} \mu \mathrm{A}$
- C4 A
- D0.4 $\mu$A
Solution
The displacement current $I_d$ in a parallel plate capacitor is defined as the current flowing through the circuit, which is equal to the conduction current $I_c$ in the connecting wires.
The relationship between the conduction current and the displacement current is given by $I_d = I_c$.
Given that the rms value of the current in the circuit is $I_{rms} = 4 \mu \text{A}$, this current represents the conduction current flowing through the wires connected to the capacitor plates.
According to the principle of continuity of current, the displacement current between the plates of the capacitor must be equal to the conduction current in the external circuit.
Therefore, $I_d = I_{rms} = 4 \mu \text{A}$.
Answer: 4 $\mu$A
The relationship between the conduction current and the displacement current is given by $I_d = I_c$.
Given that the rms value of the current in the circuit is $I_{rms} = 4 \mu \text{A}$, this current represents the conduction current flowing through the wires connected to the capacitor plates.
According to the principle of continuity of current, the displacement current between the plates of the capacitor must be equal to the conduction current in the external circuit.
Therefore, $I_d = I_{rms} = 4 \mu \text{A}$.
Answer: 4 $\mu$A
Hence, the correct answer is option: 'A'.