EMI Q15 DPP CHAMP

Question

In the ideal oscillating circuit, the capacitance of the capacitor is 25 F and has initial charge of 30 μC.The inductance of coil is 0.04 H. The maximum magnitude of current in the circuit, after closing the switch K is

Answer 1

Current will be maximum when

\(\frac 12 \frac {Q^2}{C} = \frac 12 Li^2\)

\(i^2 = \frac{Q^2}{LC}\)

\(i = \frac Q{\sqrt {LC}}\)

\(i = \frac{30\times 10^{-6}}{\sqrt{0.04 \times 25 \times 10^{-6}}}\)

\(i = \frac{30 \times 10^{-6}}{1 \times 10^{-3}}\)

\(i = 30 \times 10^{-3}\) Ampere