Find the time constant of the circuit. What is the charge of the capacitor 1.87

Question

Find the time constant of the circuit. What is the charge of the capacitor 1.87 time constants after the circuit is closed? What is the charge after a long time?

You charge an initially uncharged 86.7-mF capacitor through a 39.1-Q resistor by means of a 9.00-V battery having negligible internal resistance. Find the time constant of the circuit. What is the charge of the capacitor 1.87 time constants after the circuit is closed? What is the charge after a long time? Time constant: Number 3.38 Charge Number Charge after long time: Number

Explanation / Answer

time constant T = R*C = 39.1*86.7*10^-3 F = 3.38997 s
(86.7 milliFarad capacitor is what you've written)

Capacitor voltage V = 9.0V * (1 - e^-t/T)
At t=1.87T
V = 9.0V * (1 - e^-1.87) = 7.61288 V

Charge is then Q = C*V = 86.7*10^-3 *7.61288 C = 0.660037 C

After a long time, V = 9.0 V (full charge)
Then Q = C*V = 0.660037 C

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