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 = 8.50 V, r; = 10 kn, and r2 = 16 kn.) =10.0 µF 3.00 k2 Figure P28.67 (a) Find the steady-state current in each resistor. I = 327 I2 = 327 13-kn = 0 HA V HA HA (b) Find the charge Q on the capacitor. 52 (c) The switch is opened at t = 0. Write an equation for the current IR, in R2 as a function of time. O (327 HA)e-t/(0.190 s) O (275 µA)et/(0.190 s) O (275 µA)e-t/(0.190 s) O (327 µA)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. ms

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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 = 8.50 V, r1 = 10 kN, and r2 = 16 kN.)
10.0 µF
3.00 k2
Figure P28.67
(a) Find the steady-state current in each resistor.
I = 327
HA
I2 = 327
HA
13-kn = 0
HA
(b) Find the charge Q on the capacitor.
52
(c) The switch is opened at t = 0. Write an equation for the current IR, in R2 as a function of time.
O (327 HA)e-t/(0.190 s)
O (275 µA)et/(0.190 s)
O (275 µA)e-t/(0.190 s)
O (327 µA)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.
ms
Transcribed Image Text: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 = 8.50 V, r1 = 10 kN, and r2 = 16 kN.) 10.0 µF 3.00 k2 Figure P28.67 (a) Find the steady-state current in each resistor. I = 327 HA I2 = 327 HA 13-kn = 0 HA (b) Find the charge Q on the capacitor. 52 (c) The switch is opened at t = 0. Write an equation for the current IR, in R2 as a function of time. O (327 HA)e-t/(0.190 s) O (275 µA)et/(0.190 s) O (275 µA)e-t/(0.190 s) O (327 µA)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. ms
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