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. (8 = 8.80 v, r1 = 11 ko, and r2 = 17 kA) 3.00 ks2 Figure P28.67 (a) Find the steady-state current in each resistor. HA 3-kn (b) Find the charge Q on the capacitor HC (c) The switch is opened at t = 0, write an equation for the current l, in R2 as a function of time. O O O O (267 e-t(0.200 s) (267 et/(0.200 s) (314 es(0.200 s) (314e-t/(0,200 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

a) I1 = V/(r1+r2) = 8.8/(11+15) = 338.4 uA

I2 = 338.4 uA

I3 = 0 A

b) charge on the capacitor = C*V = 50.8 uC

c) current in r2 = (338 µA)e-t /(0.180 s) <--ans option (a)

d) time taken = 0.29 s = 289.7 ms

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