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 = 1.50 M?, V = 56.0 V, and C = 2.50 F.

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

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

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

I(t) =   

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 = 1.50 M?, V = 56.0 V, and C = 2.50 F. (a) What is the battery current just after switch S is closed? I0 = A (b) What is the battery current a long time after switch S is closed? I = A (c) The switch has been closed for a long time. At time t = 0 the switch is then opened. Find the current in A through the 600 k? resistor as a function of time. (Use the following as necessary: t.) I(t) =

Explanation / Answer

a)

Just after switch is closed , capacitor is short circuited

Using Ohm's law

I = 56/(1.5)

I = 37.33 uA

b)

after a long time ,

capacitor is open circuited

Using Ohm's law

I = 56/(1.5 + .600)

I = 26.67 uA

c)

Here ,

time constant , T = 2.5 *10^-6 * 600*10^3

T = 1.5 s

Now, Using Voltage Divider

Vc = .600*56/(.6 + 1.5)

Vc = 16 V

now , current in 600 K after opening the switch

I = Io*e^(-t/T) uA

I = 16/.6 e^(-t/1.5)

I(t) = 26.67 e^(-t/1.5) uA

the current in

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