Question

9. A certain capacitor stores 33 J of energy when it holds 1,507 uC of charge. What is the capacitance in nF? 10. A R-C series circuit is connected to a 12 volt battery. Initially the capacitor has zero charge. Assume R= 46 Ω, C= 1,171 μF. When the circuit is closed, calculate the time (in Millisecond) when the capacitor voltage will be 63.21% of its final voltage.

10. A R-C series circuit is connected to a 12 volt battery. Initially the capacitor has zero charge. Assume R= 46 Ω, C= 1,171 μF. When the circuit is closed, calculate the time (in Millisecond) when the capacitor voltage will be 63.21% of its final voltage.

Answer 1

9 )

U = 33 J

q = 1,507 uC

= 1,507 x 10^-6 C

U = q² / 2 C

C = q² / 2 U

= ( 1507 x 10^-6 ) / 2 x 33

= 3.4409 x 10^-8 F

= 34.409 x 10^-9 F

so the capacitance in nF is C = 34.409 nF

10 )

given

V = 12 V

R = 46 Ω

C = 1,171 μF

the capacitor voltage will be 63.21% of its final voltage

V_t = 0.6321 V_o

using equation

V_t = V_o ( 1 - e^-t/RC )

0.6321 V_o = V_o ( 1 - e^-t/RC )

0.6321 = ( 1 - e^-t/RC )  

- e^-t/RC = - 0.3679

- e^{-t/ 46 x 1171 x 10-6} = - 0.3679

- e^{-t/ 0.05386} = - 0.3679

applying ln on both sides

- 0.9999441 = - t / 0.053866

0.9999441 = t / 0.053866

t = 0.9999441 x 0.053866

t = 0.053862 sec

or

= 53.862 x 10^-3

t = 53.862 Milliseconds.