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

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) $18 \times 10^{-5}\,\text{C}$.