Calculating Current in a Circuit with Capacitors and Resistors

[vsage]

C = 7.20 micro-Farads, R1 = 10.5ohm, R2 = 22.5ohm, V0 = 16.0V. What is the current (to within 0.01A) through R1 immediately after the switch S is closed? I created two equations to represent the voltage drops in the system but Capacitors charge up over time which is confusing me:

16V - i1*R1 - i2*r2 = 0
16V - i1R1 - q/C = 0

Therefore i2*R2 = q/C subtracting the two. Unfortunately that's two unknowns. I also sort of (but not fully understand) that for a resitor and capacitor in series, q = C*Vapplied*(1-e^(-t*R*C)) so at t = 0, q = 0 and thus no voltage drop across the capacitor and therefore no current. This means that all of the current is flowing through R2 right? Therefore i2 = i1 so..

16V - I1(R1+R2) = 0
16V - I1(10.5+22.5) =0
I = 0.48A?

Am I right to apply this to this drawing? Edit 1.33 was wrong. So was 0.48. OK I'm really lost now.

 

[vsage]

New theory: Is the capacitor being initially uncharged short-circuiting the system? Then the voltage drop along R1 would be 16V meaning that 16=10.5*I, I = 1.523A. Am I any closer? Edit YESS I got it :) Maybe there's hope for me yet.

 

[wais]

Hello, thank you for your question. Let's break down the problem and see if we can come up with a solution.

First, let's draw the circuit to better visualize it:

[16V]---(R1)---(C)---(R2)---[ground]

When the switch S is closed, the capacitor will begin to charge up and the current will start to flow through the circuit. At this point, we can apply Kirchhoff's laws to come up with equations that will help us solve for the current.

Kirchhoff's Voltage Law (KVL) states that the sum of all voltages in a closed loop must equal zero. In this circuit, we have two loops: one containing the voltage source and resistor R1, and the other containing the capacitor and resistor R2.

Using KVL in the first loop, we have:

16V - i1*R1 = 0

Using KVL in the second loop, we have:

16V - i2*R2 - q/C = 0

Remember that q is the charge on the capacitor and it will vary with time. However, at the initial moment when the switch is closed, the capacitor has not yet charged up, so q = 0. Therefore, our second equation becomes:

16V - i2*R2 = 0

Now, we have two equations and two unknowns (i1 and i2). We can solve for i1 by rearranging the first equation:

i1 = 16V/R1

Substituting this value into the second equation, we get:

16V - (16V/R1)*R2 = 0

Solving for i2, we get:

i2 = 16V/R2

Plugging in the given values, we get:

i1 = 16V/10.5ohm = 1.52A
i2 = 16V/22.5ohm = 0.71A

Therefore, the current through R1 immediately after the switch is closed is 1.52A.

Your approach of using the capacitor's charge equation is also correct. However, in this case, since we are looking for the current immediately after the switch is closed, we don't need to consider the time factor. We can simply use the equation q = C*V to find the charge on the capacitor, and then use that value to calculate the