In the circuit shown k1 is closed for a long time- Then k1 is opened and k2 is closed simultaneously- If R - 20 - L - 30mH and C - 12 - F- then maximum charge on the capacitor will be

Ε = iR ⇒ 100 = i_0 × 20 So, i_0 = 5 A By Conservation of Energy, 1/2 Li^2_0 = Q^2/2C ⇒ Q = i_0 √(LC) Q = 5 ×√(30 × 10^ 3× 12 × 10^ 6) = 3×10^ 3 C ...

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Byju's Answer

Standard XI

Physics

RLC Circuit

In the circui...

Question

In the circuit shown $k_{1}$ is closed for a long time. Then $k_{1}$ is opened and $k_{2}$ is closed simultaneously. If $R = 20 Ω, L = 30 m H$ and $C = 12 μ F$ , then maximum charge on the capacitor will be

3 \times 10^{- 3} C

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5 \times 10^{- 3} C

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10 \times 10^{- 3} C

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12 \times 10^{- 3} C

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Solution

The correct option is A $3 \times 10^{- 3} C$
$ε = i R \Rightarrow 100 = i_{0} \times 20$
So, $i_{0} = 5 A$

By Conservation of Energy,
$\frac{1}{2} L i_{0}^{2} = \frac{Q^{2}}{2 C}$
$\Rightarrow Q = i_{0} \sqrt{L C}$
$Q = 5 \times \sqrt{30 \times 10^{- 3} \times 12 \times 10^{- 6}} = 3 \times 10^{- 3} C$

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In the circuit shown k1 is closed for a long time. Then k1 is opened and k2 is closed simultaneously. If R=20Ω,L=30mH and C=12μF, then maximum charge on the capacitor will be

The correct option is A 3×10−3Cε=iR⇒100=i0×20 So, i0=5A By Conservation of Energy, 12Li20=Q22C ⇒Q=i0√LC Q=5×√30×10−3×12×10−6=3×10−3C

In the circuit shown $k_{1}$ is closed for a long time. Then $k_{1}$ is opened and $k_{2}$ is closed simultaneously. If $R = 20 Ω, L = 30 m H$ and $C = 12 μ F$ , then maximum charge on the capacitor will be