# charging capacitor's capacitance

Discussion in 'Homework Help' started by lemon, May 24, 2010.

1. ### lemon Thread Starter Member

Jan 28, 2010
125
2
Hi:
I'm revising for an exam coming this Wednesday and have this question in my revision notes I cannot answer.

A circuit consists of an uncharged capacitor connected in series with a 15kΩ resistor and has a time constant of 30 ms. Determine
a) the capacitance of the capacitor and
b) the voltage drop across the capacitor
20 ms after connecting the circuit to a 100 V d.c. supply.

I can find a voltage of 51.3V after 20 ms. And can find the initial current of 6.67mA. But am not sure how to find the capacitance after this time.
I know how to find the capacitance of capacitor in series, but there is only one capacitor here.

Should I use C=Q/V? Where the voltage is 51.3V? But then what would Q be.

any help appreciated
thank you

Last edited: May 24, 2010
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
time constant = R * C = 30ms = 15K * C

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
Well you could uses this equation to.
C = Q/V = I*t / V = (6.67mA * 30ms)/ 100V = 2uF

4. ### lemon Thread Starter Member

Jan 28, 2010
125
2
ahh! I think I should have gotten that one.
Thank you Jony130