In the figure below, suppose the switch has been closed for a time interval suff
ID: 1551064 • 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 = 15.0 k Ohm, R_2 = 17.0 k Ohm, R_3 - 8.00 k Ohm, and C = 17.0 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_R2 = Find the time interval required for the charge on the capacitor to fall to one-fifth its initial value. msExplanation / Answer
(A) when C is fully charged then no current will pass through capacitor.
so I3 = 0
current will only pass through R1 and R2.
I1 = I2 = 9 / (15 + 17) = 0.28 mA = 281.25 uA
(B) VC = V_R2 = 0.28mA x 17 kohm = 4.78 Volt
Qmax = C Vmax = (17 uF ) (4.78 V ) = 81.3 uC
(C) I = Imax [ e^(-t/T)]
Imax = 4.78 / (17 + 8) = 0.1912 mA = 191.2 uA
T = (R2 + R3) C = (8 + 17) x 10^3 x 17 x 10^-6 = 0.425 s
I = 191.2 uA [ e^(-t / 0.425)]
(D) 1/5 = e^(-t / 0.425 )
- t / 0.425 = ln(1/5)
t =0.684 sec