Question

In an oscillating LC circuit, L = 3.00 mH and C=2.70µF. At t = 0 the charge on the capacitor is zero and the current is 2.00A. The maximum charge that will appear on the capacitor will be :

• A. 1.8 × 10^-5 C
• B. 18 × 10^-5 C
• C. 9 × 10^-5 C
• D. 90 × 10^-5 C

Answer 1

The Correct Option is B

Step 1: Understand Energy Conservation in LC Oscillations

In an LC circuit without resistance, the total energy is constantly exchanged between the electric field energy stored in the capacitor and the magnetic field energy stored in the inductor. The total energy remains constant.

Step 2: Write the expressions for Energy stored in Capacitor and Inductor

Energy stored in the capacitor, $U_C = \frac{1}{2} \frac{Q^2}{C}$

Energy stored in the inductor, $U_L = \frac{1}{2} L I^2$

Total energy in the LC circuit, $E = U_C + U_L = \frac{1}{2} \frac{Q^2}{C} + \frac{1}{2} L I^2$

Step 3: Use Initial Conditions to find Total Energy

At $t = 0$, charge on the capacitor $Q = 0$, and current $I = 2.00 \, \text{A}$.

$U_C (t=0) = \frac{1}{2} \frac{0^2}{C} = 0$

$U_L (t=0) = \frac{1}{2} L I^2 = \frac{1}{2} \times (3.00 \times 10^{-3} \, \text{H}) \times (2.00 \, \text{A})^2$

$U_L (t=0) = \frac{1}{2} \times 3.00 \times 10^{-3} \times 4.00 = 6.00 \times 10^{-3} \, \text{J}$

Total Energy $E = U_C (t=0) + U_L (t=0) = 0 + 6.00 \times 10^{-3} \, \text{J} = 6.00 \times 10^{-3} \, \text{J}$

Step 4: Condition for Maximum Charge on Capacitor

The maximum charge on the capacitor occurs when the current in the inductor is zero. At this point, all the energy is stored in the capacitor in the form of electric field energy.

At maximum charge $Q_{max}$, current $I = 0$.

$U_C (Q_{max}) = \frac{1}{2} \frac{Q_{max}^2}{C}$

$U_L (I=0) = \frac{1}{2} L (0)^2 = 0$

Total Energy $E = U_C (Q_{max}) + U_L (I=0) = \frac{1}{2} \frac{Q_{max}^2}{C} + 0 = \frac{1}{2} \frac{Q_{max}^2}{C}$

Step 5: Equate Total Energy and Solve for Maximum Charge

Since total energy is conserved:

$6.00 \times 10^{-3} \, \text{J} = \frac{1}{2} \frac{Q_{max}^2}{C}$

$Q_{max}^2 = 2 \times E \times C = 2 \times (6.00 \times 10^{-3} \, \text{J}) \times (2.70 \times 10^{-6} \, \text{F})$

$Q_{max}^2 = 12.00 \times 2.70 \times 10^{-9} = 32.4 \times 10^{-9} = 3.24 \times 10^{-8}$

$Q_{max} = \sqrt{3.24 \times 10^{-8}} = \sqrt{3.24} \times \sqrt{10^{-8}} = 1.8 \times 10^{-4} \, \text{C}$

$Q_{max} = 1.8 \times 10^{-4} \, \text{C} = 18 \times 10^{-5} \, \text{C}$

Step 6: Choose the correct option

The calculated maximum charge is $18 \times 10^{-5} \, \text{C}$, which matches option (B).

Final Answer: The final answer is ${18 \times 10^{-5} \, \text{C}}$

Answer 2

In an LC circuit, the energy oscillates between the electric field energy stored in the capacitor and the magnetic field energy stored in the inductor. When the charge on the capacitor is zero, the current is maximum, and all the energy is stored in the inductor. The energy stored in the inductor is given by:

$U_L = \frac{1}{2}LI^2$

When the current is zero, the charge on the capacitor is maximum, and all the energy is stored in the capacitor. The energy stored in the capacitor is given by:

$U_C = \frac{1}{2}\frac{Q_\text{max}^2}{C}$

By conservation of energy, the maximum energy stored in the inductor is equal to the maximum energy stored in the capacitor:

$\frac{1}{2}LI^2 = \frac{1}{2}\frac{Q_\text{max}^2}{C}$

$Q_\text{max} = I\sqrt{LC}$

Substituting the given values:

$Q_\text{max} = (2.00\,\text{A})\sqrt{(3 \times 10^{-3}\,\text{H})(2.70 \times 10^{-6}\,\text{F})} = 2\sqrt{8.1 \times 10^{-9}} = 2(9 \times 10^{-5}) = 18 \times 10^{-5}\,\text{C}$

The correct answer is (B) $18 \times 10^{-5}\,\text{C}$.