Finding the Resistance in an RC Circuit

[hamhamt]

Homework Statement

A 10-µF capacitor is charged by a 10-V battery through a resistance R. The capacitor reaches a potential difference of 4 V in a period of 3 s after the charging began. Find the value of R.

Homework Equations

c=q/v
q=CE(1-e^-t/RC)
i=(E/R)e^(-t/RC)

The Attempt at a Solution

I found the charge on the capacitor when its at 4 volts and got 4.0 * 10^-5 C. I then tried to plug this into the equation for a capacitor charging, but I don't know the time at which the capactior is at 4 volts.

At this point, I attempted finding another equation, so I could attempt a system of 2 equations with 2 unknowns. I tried to use the current in a capactor charging equation, but I ended up getting infinitely many solutions.

I am not entirely sure how to approach this problem.

 

[wukunlin]

I don't know the time at which the capactior is at 4 volts.

I thought...

The capacitor reaches a potential difference of 4 V in a period of 3 s after the charging began.

 

[XianForce]

Use c = q/v to find the charge. Then plug that into the second equation at t = 3 s, because the capacitor begins charging at t = 0. So since it reaches 4 V after 3 s of charging, t = 3 s when the capacitor is charged to 4 V.

 

[hamhamt]

Jesus. I had half of the question cut off at the most important part. Thank you both lol

 

[Nickie]

I would approach this problem by first understanding the basic principles of an RC circuit. An RC circuit consists of a resistor (R) and a capacitor (C) connected in series with a voltage source (E). When the circuit is closed, the capacitor starts to charge up and the current decreases over time until it reaches a steady state.

To find the value of R in this circuit, we can use the equation q=CE(1-e^-t/RC), where q is the charge on the capacitor, C is the capacitance, E is the voltage of the battery, t is the time, and R is the resistance. We know the capacitance (C=10 µF), the voltage of the battery (E=10 V), and the time it takes for the capacitor to reach a potential difference of 4 V (t=3 s).

Plugging in these values into the equation, we get:

4.0*10^-5 C = (10*10^-6 F)(10 V)(1-e^-3/R(10*10^-6 F))

Solving for R, we get R= 7.5 kΩ.

Therefore, the value of R in this circuit is 7.5 kΩ.