Question

A capacitor of capacity 2 µF is charged to a potential difference of 12 V. It is then connected across an inductor of inductance 0.6 mH. The current in the circuit at a time when the potential difference across the capacitor is 6.0 V is ______ × 10^-1A.

Options

• 2

• 4

• 6

• 8

Answer 1

A capacitor of capacity 2 µF is charged to a potential difference of 12 V. It is then connected across an inductor of inductance 0.6 mH. The current in the circuit at a time when the potential difference across the capacitor is 6.0 V is 6 × 10^-1A.

Explanation:

Given,

Capacitance of the capacitor (C) = 2µF

= 2 × 10^–6F

Initial potential difference across capacitor

(V₀) = 12 V

Inductance of the inductor (L) = 0.6 mH

= 0.6 × 10^–3 H

= 6 × 10^–4 H

Now potential difference across the capacitor (V) = 6.0 V

Current in the circuit (I) =?

Initially,

Finally,

Now, applying the conservation of energy

`1/2"CV"^2=1/2"LI"^2+1/2"CV"^2`

Now, substituting the values

`1/2xx2xx10^-6xx12^2`

= `1/2xx6xx10^-4xx"I"^2+1/2xx2xx10^-6xx6^2`

⇒ 144 × 10^–6 J

⇒ = 3I² × 10^–4 + 36 × 10^–6 J

⇒ 3I² × 10^–4 = 144 × 10^–6J – 36 × 10^–6J

⇒ I² = `(108xx10^-6"J")/(3xx10^-4)` 

⇒ I² = 36 × 10^–2

⇒ I = `sqrt0.36`

⇒ I = 0.6 A

⇒ I = 6 × 10^–1 A