In a combination of an inductor of 1.6 mH and a capacitor connected in parallel, a maximum energy of \( 72\times10^{-6} \) J oscillates between inductor and capacitor. The peak current will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.3 A
In LC oscillations the total energy is conserved. When the current is at its peak the capacitor is uncharged, so the whole energy is stored in the magnetic field of the inductor.
\( \displaystyle U = \frac{1}{2} L I_0^{2} \)
\( \displaystyle I_0 = \sqrt{\frac{2U}{L}} = \sqrt{\frac{2 \times 72\times10^{-6}\,\mathrm{J}}{1.6\times10^{-3}\,\mathrm{H}}} \)
\( \displaystyle I_0 = \sqrt{0.09}\,\mathrm{A} = 0.3\,\mathrm{A} \)