A 320uF capacitor is charged using a 10V battery. If it is then discharged by connecting it to a coil of negligible resistance and inductance 2 mH , the peak value of the current in the circuit is

To find the peak current in the circuit when a charged capacitor discharges through an inductor, we can use some fundamental principles of electrical circuits. Let's break this down step by step.

Understanding the Circuit Components

In this scenario, we have a capacitor with a capacitance of 320 microfarads (µF) charged to a voltage of 10 volts (V). When this capacitor discharges through an inductor with an inductance of 2 millihenries (mH), we can analyze the behavior of the circuit using the formula for the peak current in an LC circuit.

Key Formulas

The peak current (\(I_{peak}\)) in an LC circuit can be calculated using the formula:

I_{peak} = \frac{V}{R}

However, since we have negligible resistance in this case, we can also use the energy stored in the capacitor to find the peak current. The energy (\(E\)) stored in the capacitor is given by:

E = \frac{1}{2} C V^2

Where:

• C = capacitance in farads (F)
• V = voltage in volts (V)

Calculating Energy Stored in the Capacitor

First, we need to convert the capacitance from microfarads to farads:

C = 320 µF = 320 \times 10^{-6} F = 0.00032 F

Now, we can calculate the energy stored in the capacitor:

E = \frac{1}{2} \times 0.00032 \times (10)^2

E = \frac{1}{2} \times 0.00032 \times 100 = 0.016 Joules

Using Energy to Find Peak Current

In an LC circuit, the energy stored in the capacitor is transferred to the inductor. The peak current can be found using the relationship between energy and current:

E = \frac{1}{2} L I_{peak}^2

Where:

• L = inductance in henries (H)
• I_{peak} = peak current in amperes (A)

We need to convert the inductance from millihenries to henries:

L = 2 mH = 2 \times 10^{-3} H = 0.002 H

Now, we can rearrange the energy formula to solve for \(I_{peak}\):

I_{peak} = \sqrt{\frac{2E}{L}}

Substituting the values we have:

I_{peak} = \sqrt{\frac{2 \times 0.016}{0.002}}

I_{peak} = \sqrt{\frac{0.032}{0.002}} = \sqrt{16} = 4 A

Final Result

The peak value of the current in the circuit when the capacitor discharges through the inductor is 4 amperes. This demonstrates how energy is conserved in the system, transitioning from the electric field of the capacitor to the magnetic field of the inductor.