Solving 2 Loop RC Circuit: Find I1(0)

[kjlchem]

Homework Statement

A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 49 Ω, R3 = 64 Ω and R4 = 64 Ω. The capacitance is C = 88 μF and the battery voltage is V = 24 V.

The switch has been open for a long time when at time t = 0, the switch is closed. What is I1(0), the magnitude of the current through the resistor R1 just after the switch is closed?

Homework Equations

C = Q/V
V=IR

The Attempt at a Solution

Currents in series are the same, so I2=I3. Currents going into a junction = currents coming out so, I1=I4. I think it might all be the same current.

V= I1(R1+R4) + Q/C
Q/C=(R2+R3)I2

I don't know how to find Q/C or I2.

 

[kjlchem]

Forgot to attach the diagram!

 

[gneill]

Immediately before the switch is closed, what's the voltage (and charge) on the capacitor?
What then is the voltage on the capacitor immediately after the switch is closed?

 

[kjlchem]

The voltage on the capacitor before the switch is closed is 0, I believe . After it is = To the v at the battery

 

[gneill]

The voltage on the capacitor before the switch is closed is 0, I believe . After it is = To the v at the battery

Not quite. It is zero immediately before and therefore immediately after the switch is closed, too.

Capacitors have a sort of inertia... they won't let their potential differences change instantaneously (much as inductors won't allow the current flowing through them to change instantaneously).

After some long period of time the voltage on the capacitor will eventually reach some steady state value as determined by the resistor network surrounding it.

 

[kjlchem]

Cool! I did not know that. Thank you!