ID webassign.net 0 Gocgle School EmailmyUK CANVAS RandomSocial Watch ESPN Live (

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ID webassign.net 0 Gocgle School EmailmyUK CANVAS RandomSocial Watch ESPN Live (c) What is the value of the final steady-state current? (d) How long does it take the current to reach 80% of its maximum value? 1.88 ms Need Help? Talk to a Tutor 2. 0/2 points Previous Answers My Notes = 8.0 V and L = 1.2 H. The switch in the figure below is open for t 0 and then dosed at time t-0. Find the current in the inductor and the current in the switch as functions of time thereafter. Let (Use the following as necessary: t in seconds.) 4.00 8.00 8 4.00 Inductor- Iswitch 1.2 -0.2e Submit Answer Save Progress 3. 3/3 points | Previous Answers My Notes A 140 mH inductor and a 4.80 resistor are connected with a switch to a 6.00 V battery as shown in Figure P32.27

Explanation / Answer

2. for the given circuit the switch is closed at t= 0

so let after time t

current through the inductor be i and currnet through the battery be I

then from kirchoff's loop law

E = 4I + 4(I - i)

E = 4I + 8i + Ldi/dt

now E = 8 V

L = 1.2 H

then

2 = I + I - i = 2I - i

8 = 4I + 8i + 1.2di/dt

8 = 4 + 2i + 1.2di/dt

1.2di/dt = 4 - 2i

0.6di/dt = 2 - i

di/(2 - i) = dt/0.6

integrating

0.6ln(2 - i) = -t + k [ where k = constant]

now when t = 0, as soon as the switch is closed

the inductor acts as open switch and hence i = 0

so

0.6ln(2) = k

hence

ln(1 - i/2) = -t/0.6

1 - i/2 = e^(-t/0.6)

i = 2(1 - e^(-t/0.6))

and curent through switch I = 1 - i/2 = e^(-t/0.6)

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