a) the charge on the capacitor... ...a long time after the switch is closed q =

Question

a) the charge on the capacitor...
...a long time after the switch is closed
q = ?

...4.1 s after the switch is closed
q = ?


(b) the current in the resistor...
...immediately after the switch is closed
I0 = ?

...4.1 s after the switch is closed
I = ?

... a long time after the switch is closed
I = ?


(c) The rate at which energy, 4.1 s after the switch is closed, is...
...being dissipated in the resistor
Presistor = ?

...being stored in the capacitor
Pcapacitor = ?

...being delivered by the battery
Pbattery = ?

Explanation / Answer

the charge on the capacitor q = C*E*(1-e^-(t/RC))

potential across capacitor Vc = E*(1-e^-(t/RC))

current in the resistor changes i = (E/R)*e^-(t/RC)

(a)

at time t = infinity

q = C*E = 1.3*10 = 13 uC

t = 4.1

q = 1.3*10*(1-e^-(4.1/(2.2*10^6*1.3*10^-6)))

q = 9.9 uC

=====================

(b)

Io = (E/R)*e^-0 = E/R

Io = 10/(2.2*10^6) = 4.54*10^-6 A

time t= 4.1 s

I = (10/(2.2*10^6))*(e^-(4.1/(2.2*10^6*1.3*10^-6)))

I = 1.08*10^-6 A

time t = infinity

I = 0

==========================

The rate at which energy, 4.1 s after the switch is closed, is being dissipated in the resistor

Presistor = I^2*R = (1.08*10^-6)^2*2.2*10^6 = 2.6*10^-6 W

potential across capacitor Vc = 10*e^-(4.1/(2.2*10^6*1.3*10^-6)) = 7.61 v

The rate at which energy, 4.1 s after the switch is closed, is being stored in the capacitor

Pcapacitor = Vc*I = 7.61*1.08*10^-6 = 8.2*10^-6 W

The rate at which energy, 4.1 s after the switch is closed, is being delivered by the battery

Pbattery = E*I = 10*1.08*10^-6 = 1.08*10^-5 W

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