In the figure below, suppose the switch has been closed for a time interval suff
ID: 1357523 • Letter: I
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 = 14.0 k Ohm, R_2 = 22.0 k Ohm, R_3 = 5.00 k Ohm, and C = 17.0 mu F.) Find the steady-state current in each resistor. Find the charge Q_max on the capacitor. 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.) Find the time interval required for the charge on the capacitor to fall to one-fifth its initial value.Explanation / Answer
a)
at steady state , capacitor behaves as open circuit , hence the circuit only consists of battery voltage and R1 and R2 in series
Total resistance = Rtotal = R1 + R2 = 14000 + 22000 = 36000 ohm
total current flowing = itotal = V/Rtotal = 9 / 36000 = 2.5 x 10-4 A
I1 = current in R1 = itotal = 2.5 x 10-4 A = 250 uA
I2 = current in R2 = itotal = 2.5 x 10-4 A = 250 uA
I3 = current in R3 = 0
b)
Voltage across R2 = V2 = i2 R2 = 2.5 x 10-4 x 22000 = 5.5 volts
Voltage across capacitor = V = Voltage across R2 = 5.5 volts
charge stored in capacitor = Qmax = CV = 17 x 10-6 x 5.5 = 93.5 uC
c)
time constant , T = (R1 + R2) C = (14 + 22) x 103 x 17 x 10-6 = 0.612 sec
charge flowing from the capacitor at any time is given as
q = Qmax e-t/T
taking derivative both side relative to "t"
i = Qmax (-1/T) e-t/T
d)
q = Qmax e-t/T
Qmax /5 = Qmax e-t/T
e-t/0.612 = 0.2
(-t/0.612) = -1.61
t = 0.985 sec
t = 985 ms