Capacitance Question 5
Question 5 - 2024 (29 Jan Shift 1)
A capacitor of capacitance $100 \mu \mathrm{F}$ is charged to a potential of $12 \mathrm{~V}$ and connected to a $6.4 \mathrm{mH}$ inductor to produce oscillations. The maximum current in the circuit would be :
(1) $3.2 \mathrm{~A}$
(2) $1.5 \mathrm{~A}$
(3) $2.0 \mathrm{~A}$
(4) $1.2 \mathrm{~A}$
Show Answer
Answer: (2)
Solution:
By energy conservation
$\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \mathrm{LI}_{\max }^{2}$
$\mathrm{I}_{\max }=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \mathrm{V}$
$=\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} \times 12$
$=\frac{12}{8}=\frac{3}{2}=1.5 \mathrm{~A}$
