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 R_1 = 11.0 k Ohm, R_2 = 17.0 k Ohm, R_3 = 5.00 k Ohm, and C = 13.0 mu F.) Find the steady-state current in each resistor. I_1 = mu A I_2 = mu A I_3 = mu A Find the charge Q_max on the capacitor. mu C The switch is now opened at t = 0. Write an equation for the current in R_2 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.) I_R_2 = Find the time interval required for the charge on the capacitor to fall to one-fifth its initial value. ms

Explanation / Answer

Here , for the steady state current

a)

Rnet = 11 + 17 = 28 kOhm

I1 = 9/(28 *10^3) = 321.44 uA

I2 = I1 = 321.44 uA

I3 = 0 V

b)

for the max charge

maximum charge = 9 * (17/(17 + 11)) *( 13)

maximum charge = 71 uC

c)

when the switch is opened

T = (5 *10^3 + 17 *10^3) * 13 *10^-6

T = 0.286 s

for the current

I = (71/(13 * (17 + 5))) * e^(-t/.286)

I = 0.248 * e^(-t/.286) mA

d)

let the time needed is t

0.248/5 = 0.248 * e^(-t/.286)

solving for t

t = 0.461 s = 461 ms

the time taken is 461 ms

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