A circuit is constructed with four resistors, one inductor, one battery and a sw

Question

A circuit is constructed with four resistors, one inductor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 23 , R3 = 109 and R4 = 124 . The inductance is L = 238 mH and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.

1)

The switch has been open for a long time when at time t = 0, the switch is closed. What is I4(0), the magnitude of the current through the resistor R4 just after the switch is closed?

A

Your submissions:

2)

What is I4(), the magnitude of the current through the resistor R4 after the switch has been closed for a very long time?

A

Your submissions:

3)

What is IL(), the magnitude of the current through the inductor after the switch has been closed for a very long time?

A

Your submissions:

4)

After the switch has been closed for a very long time, it is then opened. What is I3(topen), the current through the resistor R3 at a time topen = 3.8 ms after the switch was opened? The positive direction for the current is indicated in the figure.

A

Your submissions:

5)

What is VL,max(closed), the magnitude of the maximum voltage across the inductor during the time when the switch is closed?

V

Your submissions:

6)

What is VL,max(open), the magnitude of the maximum voltage across the inductor during the time when the switch is open?

Explanation / Answer

Given,

R1 = R2 = 23 , R3 = 109 and R4 = 124

L = 238 mH ; V = 12 V

1)The equivalent resistance is:

branch with L gets sort out.

Req = R1 + R3 + R4

Req = 23 + 109 + 124 = 256

I = 12/256 = 0.047 A

same current flows through the resistors in series. so:

Hence I4 = 0.047 A

2)Now L acts like an infinite resistor

Req = R4 + R1 + R23

R23 = 23 x 109/(109 + 23) = 18.99

Req = 124 + 23 + 18.99 = 165.99 Ohm

I = 12/165.99 = 0.072 A

Hence, I4 = 0.072 A

3)V = IR

V23 = 0.072 x 18.99 = 1.367 V

I = V2/R2 = 1.367/23 = 0.059 A

Hence, I = 0.059 A

4)t = 3.8 x 10^-3

we know that

I = I e^-t/tau

Tau = L/R = L/(R3 + R2) = 0.238/(23 + 109) = 0.0018

I = 0.059 e^-(3.8 x 10^-3/1.8 x 10^-3) = 0.059 x 0.121 = 0.0071 A

Hence, I = -0.0071 A

[As per the guidelines, first four parts has been done. Kindly put other separately. Thanks. Rate it and let me know for any query]

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.