C Exercise 16.42 Part A A 1.30-F capacitor is connected to a 12.0-V battery for

Question

C Exercise 16.42 Part A A 1.30-F capacitor is connected to a 12.0-V battery for a long time, and then is disconnected The capacitor briefly runs a 2.00-W toy motor for 23.0 s After this time, by how much has the energy stored in the capacitor decreased? Submit My Answers Give Up Part B What is the voltage across the plates? Submit My Answers Give Up Part C How much charge is stored on the capacitor? Submit My Answers Give Up Part D How much longer could the capacitor run the motor, assuming the motor ran at full power until the end? Submit My Answers Give Up

Explanation / Answer

After a "long time", V_C = 12V.

Energy stored in capacitor.

U = 1/2 *Q^2 / C = 1/2 *Q*V = 1/2*C*V^2

U = 1/2 *C *V^2 = 1/2 * 1.30F * (12.0V)^2 = 93.6 J

Energy used from capacitor:

1 W = 1 J/s

Energy used = 2 J/s *23s = 46 J

Energy left:

U = 93.6 J - 46J = 47.6 J

Which equates to a voltage of:

U = 1/2* C*V^2

V = ( 2 * U / C) = ( 2 * 47.6J / 1.30F) = 8.55 V

Charge on capacitor:

C = Q / V

Q = C*V = 1.30F * 8.55V = 11.12 C

So time motor could run would be:

t = 93.6J / 2J/s = 46.8 s

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