# RC Circuits and time constants

predentalgirl1
RC Circuits!!

A capacitor in a single-loop RC circuit is charged to 63 % of its final voltage in 1.5 s. Find (a) the time constant for the circuit and (b) the percentage of the circuit's final voltage after 3.5 s.

Given that V/Vo = 63%, t = 1.5 sec , Time constant =τ
Now,
V/Vo =63%
We have, V/Vo = 1 - e –t/ τ
0.63 = 1 - e –t/ τ
e –t/ τ = 1-0.63
e t/ τ = 1/0.37
t/ τ = log 1/ 0.37
= 2.3026 x log 2.7027
1.5 /τ =2.3026 x log 2.7027
τ = 0.77???

Now to find the percentage of the circuit’s final
voltage after 3.5 sec,
V/Vo = 1- e^-3.5/.77
3.5/.77 = 2.3026 log Vo/ (Vo – V)
Or Vo/ (Vo – V) = antilog of 3.5/(.77 x 2.3026)
= 0.0106
1- (V/Vo) = 0.0106
Or V/Vo = 0.98
Therefore the percentage of the circuit’s final voltage
after 3.5 sec is 98%???

## Answers and Replies

Mindscrape
Yeah, I don't see any immediate discrepancies, and you know that 5 time constants give approximately 98% of the initial voltage, which yours does. If you made a mistake is one I can't see, but you put in numbers way too soon for me to really tell what you did. Physics rule: algebra first, numbers later.