A capacitor of capacitance 100 μF is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations.

Question

A capacitor of capacitance \(100\, \mu F\) is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be : 

(1) 3.2 A 

(2) 1.5 A 

(3) 2.0 A 

(4) 1.2 A

Answer 1

The correct option is (2) 1.5 A.

By energy conservation

\(\frac{1}{2} CV^2 = \frac{1}{2} LI_{max}^2\)

\(I_{max} = \sqrt\frac{C}{L} V\)

\(= \sqrt {\cfrac{100\times10^{-6}}{6.4 \times 10^{-3}}} \times 12\)

\(=\frac{12}{8} = \frac{3}{2} = 1.5 A\)