Capacitance Question 5
Question 5 - 2024 (29 Jan Shift 1)
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) $1.2 A$
Show Answer
Answer: (2)
Solution:
By energy conservation
$\frac{1}{2} CV^{2}=\frac{1}{2} LI _{\max }^{2}$
$I _{\max }=\sqrt{\frac{C}{L}} V$
$=\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} \times 12$
$=\frac{12}{8}=\frac{3}{2}=1.5 A$
