**SHOW ALL WORK AND EXPLANATION!!!! In the circuit in the figure, after the swit

Question

**SHOW ALL WORK AND EXPLANATION!!!!

In the circuit in the figure, after the switch has been closed for a long time, it is opened. Find the current I1 I2 and I3 immediately after switch S is opened. For the circuit in the figure find the current I1 I2 and I3 immediately after switch S is closed. A coil of L= 5.0 mH and a resistance of 15 Ohm are placed in series with a battery of 12 V. What is the current after 10mus? 0.8 A 0.324 A 0.59 A 0.207 A A solenoid has length 10 cm, area 5 cm2, and 100 turns. At what rate must the current in the solenoid change to produce an emf of 20 V? 3.18 Times105 A/s 3.14 Times10-6 A/s 31.84 A/s 31.45 A/s

Explanation / Answer

1.

After long time switch is closed inductor acts as short circuit

Req =10+(20*20/20+20)

Req =20 ohms

Total current

I=V/Req =150/20 =3.75 A

Current through each resistor

I1=I=7.5 A

I2=I3=7.5/2 =3.75 A

After switch is reopened ,inductor does not allow sudden change ,so

I1=0 A

I3=-I2=3.75 A

so option D is correct

2.

Immediately after switch is closed ,inductor acts as open circuit ,so Total current is

I=V/(R1+R2) =150/(10+20) =5 A

so

I1 =I2=5 A

I3=0 A

so option A is correct .

3.

Maximum current

Io=V/R =12/15 =0.8 A

Time constant

T=L/R =(5*10^-3)/15 =3.33*10^-4 s=333.33 us

In a RL circuit for Current as a function of time is given by

I=Io*e^(-t/T)

I=0.8*e^(-100/333.33)

I=0.59 A

so option C is correct

4.

Inductance of a solenoid

L=uo*N^2*A/l

L=(4pi*10^-7)*100^2*(5*10^-4)/0.1

L=6.28*10^-5 H

Induced emf

E=L(dI/dt)

so

dI/dt =E/L =20/6.28*10^-5

dI/dt=3.18*10^5 A/s

so option A is correct

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