a) What is the steady-state current through the battery? b) What is the steady-s

Question

a) What is the steady-state current through the battery?

b) What is the steady-state voltage across the 15 kOhm Resistor?

c) What is the steady-state voltage across the 10.0 microFarad Capcitor?

d) Suppose the switch is opened after the capacitor has been fully charged. What is the resulting time constant for the discharging capcitor?

Please show all work and explanation. Thanks!

) Referring to the diagram, assume that the switch Sis dosed at t=0s 120 in own 10.0 uF R=15.0 to 9.00 V 3.00 k at is the steady-state current through the battery?

Explanation / Answer

a) after long time capacitor will be fully charged.
and no current will pass through capacitor.
so current will flow only in left side loop:

I = 9 / (12 + 15) = 0.33 mA

current through battery = 0.33 mA

b) current through 15 k ohm = 0.33 mA

c) voltage across parallel branch will be same.

hence,

15I = Vc + (3 x 0)

15 x 0.33 = Vc

Vc = 5 volt

d) when switch is open then only complete circuite is right side loop.

now left loop is broken.

time constant = RC

R = 15 + 3 = 18 kohm Or 18 x 10^3 ohm

C = 10 x 10^-6 F

time constant = 18 x 10^3 x 10 x 10^-6 = 0.18 s

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