Consider the circuit shown below The symbol on the bottom is

Consider the circuit shown below. The symbol on the bottom is a switch.

Part A

Immediately after the switch is closed, what is the current that flows through the capacitor?

Part B

What is the current through R2  immediately after the switch is closed?

Part C

What is the current through R1  immediately after the switch is closed?

Part D

What is the voltage drop across R1  immediately after the switch is closed?

Part E

After a long period of time, what is the current through the capacitor?

Part F

After a long time, what is the current through R2 ?

Part G

After a long time, what is the current through R1 ?

Part H

After a long time, what is the voltage drop over R2 ?

Part I

After a long time, what is the voltage drop across R1 ?

Part J

After a long time, what is the potential drop across the capacitor?

Part K

After a long time, what is the charge on the capacitor?

Part L

Now, after the capacitor is fully charged, the switch is opened. What is the current that flows through R2  immediately after the switch is opened?

Part M

What is the time constant for the circuit after the switch is opened?

iC(t=0) =

Solution

a)

Immediately after switch is closed ,capacitor will act as short circuit ,so curren flowing through capacitor is

I_C=V/R₁

b)

Current through R₂ is zero ,since current flows through least resistive path.

I₂=0

c)

Current through R₁ is

I₁=I_c=V/R₁

d)

Voltage drop across R1 is

V₁=IR₁ = (V/R₁)R₁

V₁=V

e)

after long time switch is closed ,capacitor is fully charged and will act as open circuit ,so no current flows through capacitor ,so

I_c=0

f)

CUrrent throug R2 is

I₂=V/(R₁+R₂)

g)

Current through R1 is

I₁=I₂=V/(R₁+R₂)

h)

Voltage drop across R2 is

V₂=V*(R₂/R₁+R₁)

i)

Voltage drop across R1 is

V₁=V*(R₁/R₁+R₁)

j)

Voltage drop across the capacitor is

V_c=V*(R₁/R₁+R₁)

k)

Charge on the capacitor

Q=CV_c =C[V*(R₁/R₁+R₁)]

L)

After switch is opened

I₂=V/R₂

M)

Time Constant

T=R₂C