Question

A 2.0 uf capacitor in a circuit in series with a resistance of 1.0 MOhms is charged with a 6.0 V battery. How long would it take to charge the capacitor to 3/4 of its maximum voltage?

Answer 1

t = 2.773 s

vo:output voltage = 6*(1-e^(-t/2 sec)) = 0.75*6

1-e^(-t/2 sec) = 3/4
e^(-t/2 sec)=1/4
-t/2sec = -ln(4)
t = 2 *ln (4) = 2,773 sec

Answer 2

well we can do this as follow
t = 2.773 s
vo:output voltage = 6*(1-e^(-t/2 sec)) = 0.75*6
1-e^(-t/2 sec) = 3/4
e^(-t/2 sec)=1/4
-t/2sec = -ln(4)
t = 2 *ln (4) = 2,773 sec

Answer 3

t = 2.773 s

vo:output voltage = 6*(1-e^(-t/2 sec)) = 0.75*6

1-e^(-t/2 sec) = 3/4
e^(-t/2 sec)=1/4
-t/2sec = -ln(4)
t = 2 *ln (4) = 2,773 sec