In Figure P28.67, suppose the switch has been closed for a length of time suffic

Question

In Figure P28.67, suppose the switch has been closed for a length of time sufficiently long for the capacitor to become fully charged. (E = 9.50 r1-10 kn, and r: 15 kn.) 10.0 F 3.00 k2 Figure P28.67 (a) Find the steady-state current in each resistor 13-kn = (b) Find the charge Q on the capacitor. HC (c) The switch is opened at t = 0. Write an equation for the current IR2 in R2 as a function of me (380 AJe 0.180s) (317 le_0.180 s) (d) Find the time that it takes for the charge on the capacitor to fall to one-fifth its initial value mS

Explanation / Answer

After steady-state conditions have been reached, there is no current through the capacitor

IR3 = 0

now R1 and R2 in series

I = e/(R1+R2) = 3.8 x 10^-4 = 380 uA

in series current remain same

IR1 =380 uA

IR2 = 380 uA

Ir3 = 0 uA

part b )

Q = CV

V = potential difference at C is the same as the potential difference across R2

Q = C*I*R2 = 5.7 x 10^-5 uC = 57 uC

part c )

When the switch is opened, the branch containing R1 is no longer part of the circuit

capacitor discharge through (R2+R3)

time constant = (R2+R3)*C = 0.18 s

Io = IR2/(R2+R3) = 317 uA

I = 317 e^-t/0.180s

part d )

q = Qi*e^-t/(R2+R3)*C

Qi/5 = Qi*e^(-t/0.180s)

5 = e^t/0.180

taking natural log both side

ln5 = t/0.180

t = 290 ms

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