For the circuit shown in the figure, the switch S is initially open and the capacitor is uncharged. The switch is then dosed at time t = 0.

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Question

For the circuit shown in the figure, the switch S is initially open and the capacitor is uncharged. The switch is then dosed at time t = 0.

For the circuit shown in the figure, the switch S is initially open and the capacitor is uncharged. The switch is then dosed at time t = 0. How many seconds after closing the switch will the energy stored in live capacitor be equal to 50.2 mJ?

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Answer 1

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Answer #2

Concepts and reason

The concepts used to solve this problem are maximum charge on a capacitor in an RC circuit, charge as a function of time in an RC circuit, and energy in a charged capacitor.

Initially, use the relation between charge as a function of time and capacitance to find the energy stored in a capacitor.

Then, use the relationship between capacitance and voltage to find the maximum charge on a capacitor.

Finally, use charge as a function of time in an RC circuit to find the seconds that capacitor takes to store energy.

Fundamentals

When the capacitor joins to a battery after the switch is closed, the charge is stored in the capacitor from the battery.

The expression for charge as a function of time is

Here, time is , resistance is , capacitance is , maximum charge on the capacitor is , and the charge as a function of time is .

The expression for the energy stored in the capacitor is

The expression for maximum charge on the capacitor is

Here, the voltage is .

The expression for the energy stored in the capacitor is

Rearranging the expression in terms of charge is

Substitute for and for .

The expression for maximum charge on the capacitor is

Substitute for and for .

The expression for the charge as a function of time is

Rearranging the expression,

Take a natural log on both the sides,

Substitute for , for , for , and for .

Ans:

When the switch closes, the seconds that a capacitor takes to store energy is .

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Answer 2

Answer #1

energy stored is U = 0.5*C*V^2 = 50.2*10^-3 J

C= 90 uF

Volatge across capacitor is V = sqrt(50.2*10^-3/(0.5*90*10^-6)) = 33.4 V

then voltage across plates of capacitor is V = Vmax*1-(e^(-t/T))

Vmax = 40 V

Time constant is T = R*C = 0.5*10^6*90*10^-6 = 45 sec

V = 40*(1-e^(-t/45)) = 33.4

solving we get t = 81 sec

So the correct option is A) 81 sec

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Answer 3

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