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
Solution and Explanation
The formula for current in a capacitor within an AC circuit is:
\[ I = V \frac{\omega C}{\sqrt{1 + (\omega C)^2}} \]
The parameters are defined as follows:
- \(V = 230 \, \text{V}\) represents the supply voltage.
- \(C = 200 \times 10^{-12} \, \text{F}\) denotes the capacitance.
- \(\omega = 300 \, \text{rad/s}\) is the angular frequency.
In a capacitor, the displacement current is equivalent to the conduction current. The current flowing through the capacitor is determined by:
\[ I = V \frac{\omega C}{X_C} \]
Here, \(X_C = \frac{1}{\omega C}\) signifies the capacitive reactance.
The calculation proceeds as follows:
\[ I = \frac{230 \times 300 \times 200 \times 10^{-12}}{X_C} = 13.8 \mu\text{A} \]
Consequently, the RMS current, comprising both conduction and displacement current, is 13.8 $\mu\text{A}$.
Top Questions on Capacitors and Capacitance
- Energy stored in a capacitor is given by the equation \[ E = \frac{1}{2} C V^2 \] where: - \( C \) is the capacitance, - \( V \) is the voltage, - \( E \) is the energy stored. Given the values of \( C \), \( V \), and \( E \), determine the energy stored.}
- MHT CET - 2025
- Physics
- Capacitors and Capacitance
- Two capacitors $ C_1 = 4\mu F $ and $ C_2 = 6\mu F $ are connected in series across a 60 V battery. The potential difference across $ C_2 $ is:
- BITSAT - 2025
- Physics
- Capacitors and Capacitance
A circuit consisting of a capacitor C, a resistor of resistance R and an ideal battery of emf V, as shown in figure is known as RC series circuit.
As soon as the circuit is completed by closing key S₁ (keeping S₂ open) charges begin to flow between the capacitor plates and the battery terminals. The charge on the capacitor increases and consequently the potential difference Vc (= q/C) across the capacitor also increases with time. When this potential difference equals the potential difference across the battery, the capacitor is fully charged (Q = VC). During this process of charging, the charge q on the capacitor changes with time t as
\(q = Q[1 - e^{-t/RC}]\)
The charging current can be obtained by differentiating it and using
\(\frac{d}{dx} (e^{mx}) = me^{mx}\)
Consider the case when R = 20 kΩ, C = 500 μF and V = 10 V.- CBSE Class XII - 2025
- Physics
- Capacitors and Capacitance
- 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.
- CBSE Class XII - 2025
- Physics
- Capacitors and Capacitance
- Want to practice more? Try solving extra ecology questions todayView All Questions
