For the circuit shown in the figure, the switch S is initially open and the capacitor is...

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For the circuit shown in the figure, the switch S is initially open and the capacitor is...

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

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Answer 1

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Answer #1

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 50.2 m) 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

Q=0[1-ewt

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

Q = CV

Here, the voltage is .

The expression for the energy stored in the capacitor is

Rearranging the expression in terms of charge is

Q=2CU

Substitute for and 50.2 m) for .

0-929/P)02
=v2(90x10°F)(50.2x10*)
)
=3x108c

The expression for maximum charge on the capacitor is

Q = CV

Substitute for and for .

2. =(90ur(1059) (cov)
=(90x106F)(400)
= 3.6x10°C

The expression for the charge as a function of time is

Q=0[1-ewt

Rearranging the expression,

-0
151

Take a natural log on both the sides,

In exc = in(.)
le-
1=(RC)Inc.

Substitute 3.6x10°C for , for , 0.50 ΜΩ for , and 3x10-C for .

3.6x10-C
-=[0.50m1/ 10Msa) (90F(10° uf)(3.6–10°C)-(3x10°c))
= (0.50x10^92(90x10“F)in (6)
= 80.629s
81s

Ans:

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

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Answer 2

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