Question

A parallel plate capacitor having plate area of 50 cm$^2$ and separation of 0.1 mm is completely filled with a dielectric (dielectric constant \( K = 10 \)). The capacitor is connected to a 10 kΩ resistance and an alternating voltage \( v = 10 \sin(100\pi t) \), as shown in the figure. The switch \( S \) is initially open and then closed at \( t = 0 \). The ratio of the displacement current in the capacitor, to the current in the resistance, at time \( t = \frac{2}{\pi} \) seconds is .......... (Round off to three decimal places). 

Hint:

In an RC circuit with a capacitor, the displacement current is related to the rate of change of voltage across the capacitor, and the current in the resistance is simply \( V/R \).

Answer 1

Correct Answer: 0.036

Step 1: Write the expression for displacement current. 
The displacement current \( I_d \) in a capacitor is related to the rate of change of the charge on the capacitor. For a parallel plate capacitor, the displacement current is given by \[ I_d = C \frac{dV}{dt}, \] where \( C \) is the capacitance, and \( V \) is the voltage across the capacitor. The capacitance of a parallel plate capacitor is given by \[ C = \frac{\epsilon_0 K A}{d}, \] where \( A = 50 \, \text{cm}^2 = 50 \times 10^{-4} \, \text{m}^2 \) is the area of the plates, \( d = 0.1 \, \text{mm} = 1 \times 10^{-4} \, \text{m} \) is the separation between the plates, and \( K = 10 \) is the dielectric constant. 
Step 2: Calculate the displacement current. 
First, calculate the capacitance: \[ C = \frac{\epsilon_0 K A}{d} = \frac{(8.85 \times 10^{-12}) \times 10 \times (50 \times 10^{-4})}{1 \times 10^{-4}} = 4.425 \times 10^{-12} \, \text{F}. \] Next, find the rate of change of voltage. The voltage \( v = 10 \sin(100 \pi t) \) is given. Therefore, \[ \frac{dV}{dt} = 10 \times 100\pi \cos(100 \pi t). \] At \( t = \frac{2}{\pi} \), \[ \cos(100 \pi t) = \cos(200) = -1. \] Hence, \[ \frac{dV}{dt} = -1000 \pi \, \text{V/s}. \] Step 3: Calculate the displacement current. 
Now, substitute the values into the displacement current formula: \[ I_d = (4.425 \times 10^{-12}) \times (-1000 \pi) = -1.39 \times 10^{-8} \, \text{A}. \] Step 4: Calculate the current in the resistance. 
The voltage across the resistor is the same as the voltage across the capacitor. The current in the resistor is given by \[ I_R = \frac{V}{R} = \frac{10}{10 \times 10^3} = 1 \, \text{mA}. \] Step 5: Calculate the ratio of displacement current to the current in the resistor. 
The ratio is \[ \frac{I_d}{I_R} = \frac{-1.39 \times 10^{-8}}{1 \times 10^{-3}} = -1.39 \times 10^{-5}. \] Final Answer: The ratio of the displacement current to the current in the resistance is \( \boxed{0.014} \).