Capacitors - please, need help

[joker_900]

Capacitors - please, need urgent help!

Homework Statement

I've attached the circuit diagram but it's basically a cell connected in series with a 10kohm resistor, and then a 20microfarad capacitor and 22kohm resistor connected in parallel.

The capacitor begins charging at t=0. Show that the voltage across the capacitor as a function of time is Vc(t)=4.125(1 - exp(-t/0.1375))

Homework Equations

The Attempt at a Solution

Well i know that in a simple one resistor one capacitor series circuit, the current through the resistor is I=I0e^(-t/RC), and the voltage across the resistor is V=V0e^(-t/RC) (where V0=EMF)


Now i thought this same relationship might apply to the 10kohm (R1) resistor, as initially it's voltage = EMF, and as the capacitor charges, Vc increases and the voltage across R1 therefore decreases. However, in this situation, there is always current flow and so the voltage across R1 does not become zero.

I thought about applying kirchhoffs current law, but this did not seem to get me anywhere.


I am thoroughly confused, PLEASE HELP!

Thanks

 

[learningphysics]

I recommend posting on the image sharing site like

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attachments take a long time to approve.

 

[ogb p]

for reaching out for help with your capacitor circuit. It seems like you have a good understanding of the basic equations for resistor-capacitor circuits. In this case, you are dealing with a slightly more complex circuit with multiple resistors and a parallel connection.

To solve for the voltage across the capacitor as a function of time, you can use Kirchoff's voltage law to set up an equation relating the voltage drop across the capacitor to the voltage drop across the resistors. This can be done by considering the loop from the positive terminal of the battery, through the 10kohm resistor, the capacitor, and the 22kohm resistor, back to the negative terminal of the battery.

Using Kirchoff's voltage law, we know that the sum of the voltage drops around this loop must equal the EMF of the battery. So we can write the equation:

EMF = Vc + V1 + V2

Where Vc is the voltage across the capacitor, V1 is the voltage across the 10kohm resistor, and V2 is the voltage across the 22kohm resistor.

Next, we can use Ohm's law to express V1 and V2 in terms of the current and resistance. We know that the current through the circuit is the same everywhere, and we can use Ohm's law to express the voltage drop across each resistor as the product of the current and the resistance. So we can write:

EMF = Vc + (I*R1) + (I*R2)

Where R1 is the resistance of the 10kohm resistor, R2 is the resistance of the 22kohm resistor, and I is the current through the circuit.

Now, we can use the equation you mentioned for the current through a resistor-capacitor circuit (I=I0e^(-t/RC)) to express the current in terms of time. We can also use the fact that the voltage across the capacitor is related to the current by the equation Vc = (1/C)∫I dt, where ∫I dt is the integral of the current with respect to time.

So we can write:

EMF = Vc + (I0e^(-t/RC)*R1) + (I0e^(-t/RC)*R2)

And we can use the equation for Vc to express it in