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predentalgirl1

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**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%???

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%???