In the figure below, suppose the switch has been closed for a time interval suff

Question

In the figure below, suppose the switch has been closed for a time interval sufficiently long for the capacitor to become fully charged. (Assume R1 = 14.0 k?,R2 = 20.0 k?, R3 = 6.00 k?, and C = 14.0 ?F.)

(a) Find the steady-state current in each resistor.
I1 =  ?A
I2 =  ?A
I3 =  ?A

(b) Find the charge Qmax on the capacitor.
?C

(c) The switch is now opened at t = 0. Write an equation for the current in R2 as a function of time. (Use the following as necessary: t. Do not enter units in your answers. Assume the current is in microamperes, and t is in seconds.)

IR2 =

(d) Find the time interval required for the charge on the capacitor to fall to one-fifth its initial value.
ms

Explanation / Answer

a) in steady state a capacitor does not let direct current pass through it hence we can ssume for a steady state that the wire length having the capacitor is absent .

so , current i1 = i2 = 9/ 14+20 = 9/34 = 02647 micro amp , i3 = 0

c ) i = peak current * e ^ -t/rc , peak current = 2647 *20 / 26 = 2036.15

r = 26 , c = 14 micro f

b)charge max on the capacitor = 02647 micro amp * 20 k ohm *14 micro farad = 741.16 micro coulumbs

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