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Question: A capacitor of capacity 2$\mu$F is charged to a potential difference of 12V. It is then connected ac...

A capacitor of capacity 2μ\muF is charged to a potential difference of 12V. It is then connected across an inductor of inductance 0.6 mH. The current in the circuit at a time when the P.D. across the capacitor is 6.0 V is

Answer

0.6 A

Explanation

Solution

The problem describes an LC circuit where a charged capacitor is connected across an inductor, leading to LC oscillations. In an ideal LC circuit (without resistance), the total energy stored in the circuit remains constant. The energy oscillates between the electric field of the capacitor and the magnetic field of the inductor.

1. Initial Energy: Initially, the capacitor is charged to a potential difference V0=12V_0 = 12 V. The initial energy stored in the circuit is purely electrical energy in the capacitor: Einitial=12CV02E_{initial} = \frac{1}{2}CV_0^2

Given: Capacitance, C=2μF=2×10−6C = 2 \mu F = 2 \times 10^{-6} F Initial voltage, V0=12V_0 = 12 V

Einitial=12(2×10−6 F)(12 V)2E_{initial} = \frac{1}{2} (2 \times 10^{-6} \text{ F}) (12 \text{ V})^2 Einitial=(1×10−6)×144 JE_{initial} = (1 \times 10^{-6}) \times 144 \text{ J} Einitial=144×10−6 JE_{initial} = 144 \times 10^{-6} \text{ J}

2. Energy at a Later Time: At a later time, the potential difference across the capacitor is V=6.0V = 6.0 V. At this moment, the energy in the circuit is distributed between the capacitor and the inductor. Energy in the capacitor, EC=12CV2E_C = \frac{1}{2}CV^2 Energy in the inductor, EL=12LI2E_L = \frac{1}{2}LI^2 where II is the current in the circuit at that time.

Given: Voltage across capacitor, V=6.0V = 6.0 V Inductance, L=0.6 mH=0.6×10−3L = 0.6 \text{ mH} = 0.6 \times 10^{-3} H

EC=12(2×10−6 F)(6.0 V)2E_C = \frac{1}{2} (2 \times 10^{-6} \text{ F}) (6.0 \text{ V})^2 EC=(1×10−6)×36 JE_C = (1 \times 10^{-6}) \times 36 \text{ J} EC=36×10−6 JE_C = 36 \times 10^{-6} \text{ J}

3. Conservation of Energy: By the principle of conservation of energy, the total energy remains constant: Einitial=EC+ELE_{initial} = E_C + E_L 144×10−6 J=36×10−6 J+EL144 \times 10^{-6} \text{ J} = 36 \times 10^{-6} \text{ J} + E_L

Solving for ELE_L: EL=(144−36)×10−6 JE_L = (144 - 36) \times 10^{-6} \text{ J} EL=108×10−6 JE_L = 108 \times 10^{-6} \text{ J}

4. Calculate Current: Now, use the formula for energy stored in the inductor to find the current II: EL=12LI2E_L = \frac{1}{2}LI^2 108×10−6=12(0.6×10−3)I2108 \times 10^{-6} = \frac{1}{2} (0.6 \times 10^{-3}) I^2 108×10−6=(0.3×10−3)I2108 \times 10^{-6} = (0.3 \times 10^{-3}) I^2

Rearrange to solve for I2I^2: I2=108×10−60.3×10−3I^2 = \frac{108 \times 10^{-6}}{0.3 \times 10^{-3}} I2=1080.3×10−610−3I^2 = \frac{108}{0.3} \times \frac{10^{-6}}{10^{-3}} I2=360×10−3I^2 = 360 \times 10^{-3} I2=0.36I^2 = 0.36

Finally, take the square root to find II: I=0.36I = \sqrt{0.36} I=0.6 AI = 0.6 \text{ A}

The current in the circuit at a time when the P.D. across the capacitor is 6.0 V is 0.6 A.