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 f 12 kn, and r. 16 kn 10.0 AF 3.00 ko Figure P28.67 (a) Find the steady-state current in each resistor I 17.857314 HA I 17.857314 HA 3-kn (b) Find the charge Q on the capacitor. 1.028604288 X Your response differs from the correct answer by more than 10%. Double check your calculations HC (c) The switch is opened at t 0. WI an equation for the current I in R2 as a function of time. rite O (318 HA)e 400.190 s (268HA)e 400.190 s (268 HA)et/(0.190 s) (318 HA)et/(0.190 s) (d) Find the time that it takes for the charge on the capacitor to fall to one-fifth its initial value.

Explanation / Answer

(b)We know that, charge is given by:

Q = CV

Voltage across the capacitor is same as R2

V = I2R2 = 317.86 x 10^-6 x 16 x 10^3 = 5.09 Volts

Q = 10 x 10^-6 x 5.02 = 50.2 x 10^-6 = 50.2 uC

Hence, Q = 50.2 uC

d)we know that,

Q = Q(e^-t/RC)

Q/5 = Q x e^-t/RC

taking natural log both sides

ln(1/5) = -t/RC

t = -RC x ln(1/5) = RC x 1.61

Req = 3 x 16/(3 + 16) + 12 = 14.53

t = 14.53 x 10^3 x 10 x 10^-6 x 1.61 = 234 x 10^-3 s = 234 ms

HEnce, t = 234 ms

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