You have a basic \"RC\" circuit without an emf. Let R=8kOhm, C=250 nF, and the c

Q: You have a basic \"RC\" circuit without an emf. Let R=8kOhm, C=250 nF, and the c

1. initial charge Q = CV Q = 250 e-9 * 10 = 2.5 uC ------------------------------- intial currnt i = V/R = 10/8000 = 1.25 mA ---------------------------------------- Intial Power P = Vi = 10 *1.25 = 12.5 mWAtts ---------------------- use Q = Qi (1-e^-t./RC) 1-e^-t/RC = 0.1% = 0.001 e^-t/RC = 1-0.001 -t/RC = ln (0.999) -t/RC = -0.001 t = 0.001 * 8000 * 250 e-9 t = 2 micro secs

ID: 1319475 • Letter: Y

Question

You have a basic "RC" circuit without an emf. Let R=8kOhm, C=250 nF, and the capacitor is initially charged to a potential of 10V. At time t=0, the switch is closed and the capacitor begins to discharge.

1. what is the initial charge on the capacitor?
2. what is the initial current that flows in the circuit immediately ater the switch is closed?
3. what s the initial power dissipated by the resistor, immediately after the switch is closed?
4. At what time does the charge on the capacitor reach 0.100% of its initial value?

Explanation / Answer

1. initial charge Q = CV

Q = 250 e-9 * 10 = 2.5 uC

-------------------------------

intial currnt i = V/R = 10/8000 = 1.25 mA

----------------------------------------

Intial Power P = Vi = 10 *1.25 = 12.5 mWAtts

----------------------

use Q = Qi (1-e^-t./RC)

1-e^-t/RC = 0.1% = 0.001

e^-t/RC = 1-0.001

-t/RC = ln (0.999)

-t/RC = -0.001

t = 0.001 * 8000 * 250 e-9

t = 2 micro secs

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You have a basic \"RC\" circuit without an emf. Let R=8kOhm, C=250 nF, and the c

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