For the circuit shown in the figure below, switch S has been open for a long tim

Question

For the circuit shown in the figure below, switch S has been open for a long time. At time t = 0 the switch is then closed. Let R = 1.30 M, V = 42.0 V, and C = 2.00 µF.

(a) What is the battery current just after switch S is closed?
I0 = µA

(b) What is the battery current a long time after switch S is closed?
I = µA

(c) The switch has been closed for a long time. At time t = 0 the switch is then opened. Find the current in µA through the 600 k resistor as a function of time. (Use the following as necessary: t.)

I(t) =

Explanation / Answer

a) just after the switch is closed ,

capacitor will act as short cicuit

current in battery = V/R

current in battery = 42/(1.3)

current in battery = 32.3 uA

the current in battery is 32.3 uA

b)

after a long time ,

capacitor is open circuited

current in the ciruit = V/( R + 0.6)

current in the ciruit = 42/(1.3 + 0.6)

current in the ciruit = 22.1 uA

the current in the ciruit is 22.1 uA
c)

voltae across the capacitor is I

Vc = 22.1 * 0.6

Vc = 13.3 V

after opening the switch

time constant , T = 2 *10^-6 * 600 * 10^3

T = 1.2 s

I = (13.3/0.6) * e^(-t/1.2) uA

I = 22.1 * e^(-t/1.2) uA

the current in the circuit is 22.1 * e^(-t/1.2) uA

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