A parallel plate capacitor has a capacitance C = 200 pF. It is connected to 230 V ac supply with an angular frequency 300 rad/s. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :
- 1.38 μA and 1.38 μA
- 14.3 μA and 143 μA
- 13.8 μA and 138 μA
- 13.8 μA and 13.8 μA
The Correct Option is D
Approach Solution - 1
The current through a capacitor in an AC circuit is given by the formula:
\[ I = V \frac{\omega C}{\sqrt{1 + (\omega C)^2}} \]
where:
- \(V = 230 \, \text{V}\) is the supply voltage.
- \(C = 200 \times 10^{-12} \, \text{F}\) is the capacitance.
- \(\omega = 300 \, \text{rad/s}\) is the angular frequency.
The displacement current in the capacitor is the same as the conduction current in the circuit. The current through the capacitor is given by:
\[ I = V \frac{\omega C}{X_C} \]
where \(X_C = \frac{1}{\omega C}\) is the capacitive reactance.
Now we calculate:
\[ I = \frac{230 \times 300 \times 200 \times 10^{-12}}{X_C} = 13.8 \mu\text{A} \]
Thus, the rms value of the current is 13.8 $\mu\text{A}$ for both the conduction and displacement current.
Approach Solution -2
The problem asks for the root mean square (rms) value of the conduction current in the circuit and the displacement current within a parallel plate capacitor connected to an AC supply.
Concept Used:
1. Capacitive Reactance (\(X_C\)): The opposition offered by a capacitor to the flow of alternating current is called capacitive reactance. It is given by the formula:
\[ X_C = \frac{1}{\omega C} \]where \( \omega \) is the angular frequency of the AC supply and \( C \) is the capacitance.
2. Ohm's Law for AC Circuits: For a purely capacitive circuit, the rms value of the current (\(I_{\text{rms}}\)) is related to the rms value of the voltage (\(V_{\text{rms}}\)) and the capacitive reactance (\(X_C\)) by:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} \]3. Conduction and Displacement Current: - The current flowing through the connecting wires of the circuit is the conduction current (\(I_c\)). - The current that exists between the plates of the capacitor due to the changing electric field is the displacement current (\(I_d\)). - For an ideal capacitor in an AC circuit, the magnitude of the displacement current between the plates is exactly equal to the magnitude of the conduction current in the wires at all times. \[ I_c = I_d \]
Step-by-Step Solution:
Step 1: List the given values from the problem statement.
Capacitance, \( C = 200 \text{ pF} = 200 \times 10^{-12} \text{ F} \).
RMS voltage of the AC supply, \( V_{\text{rms}} = 230 \text{ V} \).
Angular frequency of the AC supply, \( \omega = 300 \text{ rad/s} \).
Step 2: Calculate the capacitive reactance (\(X_C\)) of the capacitor.
Using the formula \( X_C = \frac{1}{\omega C} \):
\[ X_C = \frac{1}{(300 \text{ rad/s}) \times (200 \times 10^{-12} \text{ F})} \] \[ X_C = \frac{1}{60000 \times 10^{-12}} \, \Omega = \frac{1}{6 \times 10^4 \times 10^{-12}} \, \Omega \] \[ X_C = \frac{1}{6 \times 10^{-8}} \, \Omega = \frac{10^8}{6} \, \Omega \]Step 3: Calculate the rms value of the conduction current (\(I_{c, \text{rms}}\)) using Ohm's law for AC circuits.
The rms conduction current is the rms current flowing in the circuit.
\[ I_{c, \text{rms}} = \frac{V_{\text{rms}}}{X_C} \]Substituting the values:
\[ I_{c, \text{rms}} = \frac{230 \text{ V}}{\frac{10^8}{6} \, \Omega} = \frac{230 \times 6}{10^8} \text{ A} \] \[ I_{c, \text{rms}} = \frac{1380}{10^8} \text{ A} = 1380 \times 10^{-8} \text{ A} \] \[ I_{c, \text{rms}} = 1.38 \times 10^{-5} \text{ A} = 13.8 \times 10^{-6} \text{ A} = 13.8 \, \mu\text{A} \]Step 4: Determine the rms value of the displacement current (\(I_{d, \text{rms}}\)).
For a capacitor, the displacement current between the plates is equal to the conduction current in the connecting wires. Therefore:
\[ I_{d, \text{rms}} = I_{c, \text{rms}} \] \[ I_{d, \text{rms}} = 13.8 \, \mu\text{A} \]The rms value of the conduction current is 13.8 µA, and the rms value of the displacement current is also 13.8 µA.
Therefore, the respective values are 13.8 µA and 13.8 µA.
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