Two uncharged identical capacitors A and B, each of capacitance C, and an inductor of inductance L are arranged as shown in the adjacent figure. At t = 0, the switch S₁ is closed while switch S₂ remains open. At time t = to = = √LC, switch?

Question

Answer

Arrangement of Capacitors and Inductor

The given arrangement consists of two uncharged identical capacitors A and B, each of capacitance C, and an inductor of inductance L.

Circuit Diagram

At t=0, switch S₁ is closed while switch S₂ remains open. The circuit diagram is shown below:

Analysis of Circuit

At t=0, the switch S₁ is closed while switch S₂ remains open. Therefore, the capacitors A and B are connected in series with the inductor L. The equivalent capacitance of the series combination of capacitors A and B is given by:

Ceq = C/2

The initial energy stored in the inductor is given by:

W = (1/2)L(I₀)²

where I₀ is the initial current flowing through the circuit, which is zero.

Switching at t = √LC

At time t = to = = √LC, switch S₂ is closed while switch S₁ remains closed. The circuit diagram is shown below:

After closing switch S₂, the equivalent capacitance of the circuit becomes:

Ceq = C

Therefore, the total energy stored in the circuit after closing switch S₂ is:

W = (1/2)×C×(Vf)²

where Vf is the final voltage across the capacitors A and B.

Final Voltage across Capacitors

The final voltage across the capacitors A and B can be calculated as follows:

Let Q be the charge on each capacitor after closing switch S₂.

Then, the potential difference across each capacitor is:

V = Q/C

Therefore, the voltage across the inductor at time t = to is given by:

VL = V = Q/C

Since the current flowing through the circuit is the same at all points, the voltage drop across the inductor is equal to the sum of the voltage drops across the capacitors A and B. Therefore, we have:

VL = VA + VB

Substituting the values of VL, VA, and VB, we get:

Q/C = Q/(2C) + Q/(2C)

Simplifying the above equation, we get:

Q = 2Q/2

Therefore, Q = 0

Hence, the final voltage across the capacitors A and B is zero.

Conclusion

Thus, after closing switch S₂ at time t = to, the energy stored in the inductor is transferred to the capacitors A and B, resulting in zero voltage across the capacitors A

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