The resistor in an RC circuit has a resistance of 181 Ω. (a) What is the capacitance must be used in this circ

ooooops ... I used "4ms" as TC instead of "2ms"

so the capacitor value should be 1.1uF ... mea culpa

however, the "current" question is still ok since it was computed using teh idea of "two time constants" ...

/***** method ok arithmetic-correction above *****/

(( oh yeah, presumption of "series" circuit ))

time constant (TC)

TC = R*C [R in ohms and C in farads]

C = TC / R = 4 x10^-3 / 181 = 2.2 x10^-6 = 2.2 uF

a) 2.2 uF

"time constant " is the time requuired to charge a capac to 63% of "full"

http://www.tpub.com/neets/book2/3d.htm

4ms = 2 TC

after one TC voltage across the cap = 9*.63 = 5.67 .. leaving (9-5.67) = 3.33 volts leeft

second TC = 3.33 * .63 = 2.1

5.67 + 2.1 = 7.77, leaving 1.23v "deficit" across the resistor

V=RI

I = V / R = 1.23 / 181 = 6.8 mA

b) 6.8 mA

The time constant is R*C, so for the time constant to be 2*10^-3 sec, R*C = 2*10^-3, where R is ohms and C is in farads.

C = 2*10^-3 / 181 = 1.1*10^-5 farad = 11µF

You did not specify if this is a series or parallel RC circuit. The time constant is the same, but the current-time function is different. In a series circuit, the current is given by

I = (V/R)*e^-t/tc , where tc is the time constant

in a parallel circuit,

I = V/R * (1 - e^-t/tc)

EDIT: this is the current in the resistor only. Your problem probably means a series circuit, since the overall current in the parallel circuit will depend on the internal resistance of the battery, which is not given.

Use the values for V (9v) and R (181 ohm) and tc given (2 ms) to find the current for t = 4 ms

e^-4/2 = e^-2 = 0.135

V/R = 9/181 = 0.05 A

Series circuit I = 0.05*0.135 = 0.0068 A = 6.8 mA

Parallel circuit I = 0.05*(1 - 0.135) = 0.043 A = 43 mA