# Homework Help: Time required to charge capacitor to percentage of max capacitor voltage

1. Jan 30, 2012

### rhemmin

1. The problem statement, all variables and given/known data

The switch in the circuit is opened at t = 0. Determine the time required for the capacitor to charge to 53% of its maximum voltage in milliseconds. Given that R1 is 8.9 kΩ, R2 is 3.1 kΩ, C is 5.1 μF and Vs is 6 volts. Round your answer off to two decimals.

2. Relevant equations

i(t)=I(max)*e^(-t/RC) (capacitor current as a function of time)
v(t)=V(max)(1-e^(-t/RC) (capacitor voltage as a function of time)

3. The attempt at a solution

I first determined that the maximum current in the circuit was V/(R1+R2) = 0.5 mA. This would both be the current through the capacitor at t =0+ and the current through the resistors once the capacitor had reached its max charge and acted as a open circuit.

I then determined that the max voltage through the capacitor would equal I(max)*R2(=1.55V) since the capacitor and R2 are in parallel. When the capacitor voltage levels off, it should behave like an open circuit so all of the current should flow through the loop including Vs, R1, and R2.

53% of max capacitor voltage should be .53(1.55) = .8215V.

I then used the capacitor voltage function listed in relevant equations to solve for t.

I'm not sure exactly which resistance value to use for the time constant. I've tried R1, R2, and R1+R2 without success. Could someone help explain

Thanks for the help!

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2. Jan 30, 2012

### Staff: Mentor

That's not right. When the switch first opens the charge (and hence the voltage) on the capacitor will be zero. What will be the voltage across R1 in this case? What is the current through R1 at that time?
Okay, that's the correct final voltage across the capacitor. So you know that the capacitor voltage will begin at zero and end up eventually at 1.55V. That allows you to write the equation for the voltage vs time as:
$$Vc = 1.55V \left(1 - e^{-\frac{t}{\tau}}\right)$$
where the time constant $\tau$ needs to be found. To find it you'll need to determine the equivalent resistance that the circuit presents across the points where the capacitor connects. So, remove the capacitor and replace the voltage supply with a short circuit. What is the net resistance across the capacitor connection points?

3. Jan 30, 2012

### rhemmin

Voltage across R1 should equal the voltage source because of KVL. So, voltage across R1 should be 6V. Therefore, the current through the R1 equals 6/(8.9*10^-3) = .674mA.

After removing capacitor and replacing voltage supply with short, resistance across the points where capacitor connects is equals the equivalent resistance of R1 and R2 in parallel.
This equivalent resistance is 2.30kΩ. After substituting these values in the voltage vs time equation, I get 8.85 ms which is graded correct by homework site.

Awesome. Thank you very much for your help.

4. Jan 30, 2012

### Staff: Mentor

You're welcome!

5. Jan 30, 2012

### technician

gneill
I have followed your help here..... it is great!

6. Jan 30, 2012

### Staff: Mentor

Why thank you. You do some good work, too

7. Apr 24, 2012

### boguzz

What if we have a Voltage Source (Vs) with a current limit lower that Imax?

Let's say:
Vs=100 V with max output current 5 mA
R1=3.5 kΩ
R2=20 kΩ
C=6 mF

In this case we have the max voltage over C as defined by the resistive voltage divider Vc(max) = Vs*R2/(R1+R2) ≈ 85 V

At t=0+ the voltage across R1 is 100V, so the maximum current is 100V/3.5kΩ≈28mA, more than the voltage source can provide.

What is the time to charge the capacitor in this situation?

Thank you!

8. Apr 24, 2012

### Staff: Mentor

Hi Boguzz, Welcome to Physics Forums.

Is that capacitor value in millifarads (mF) or should it be microfarads (μF)?

Does the voltage supply have any particular relationship between its terminal voltage and the current it supplies? Does it maintain E = 100V right up to I = 5 mA? What happens then? Does the supply voltage sag to some value that maintains a constant 5 mA even to the limit of a short circuit load?

What are your thoughts on how to approach the problem? (You must show an attempt at a solution before we can know how to help).

Last edited: Apr 24, 2012