# Discharging a capacitor through a resistor (1 Viewer)

### Users Who Are Viewing This Thread (Users: 0, Guests: 1)

#### callawee

In the circuit below, the time required (s) for the charge on the capacitor to drop to 47% of its value after the switch is closed is 2.83 s. If R = 1340 ohms, what is C in μF (microfarad)?

(dq/dt) + q/(RC) = 0

(.47q)/(t) = q/(RC)
t/(.47R)= C
Convert C from F to uF by multiplying by 10^6

I am getting 4993 uF, but that is the wrong answer and I'm not sure where I'm going wrong. Thanks for any help.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attachments

• 895 bytes Views: 164

#### LowlyPion

Homework Helper
In the circuit below, the time required (s) for the charge on the capacitor to drop to 47% of its value after the switch is closed is 2.83 s. If R = 1340 ohms, what is C in μF (microfarad)?

(dq/dt) + q/(RC) = 0

(.47q)/(t) = q/(RC)
t/(.47R)= C
Convert C from F to uF by multiplying by 10^6

I am getting 4993 uF, but that is the wrong answer and I'm not sure where I'm going wrong. Thanks for any help.
Welcome to PF.

Maybe try starting with an equation that describes the discharging of a capacitor?

Q = C*Vo*e-t/rc

Plugging the values and dividing by Q = CVo at t=0

.47 = e-2.83/1340*C

#### callawee

Thank you, the equation worked!

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving