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 = 35 , R3 = 104 and R4 = 87 . The inductance is L = 283 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

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

3)

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

_____________A

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 = 3.3 ms after the switch was opened? The positive direction for the current is indicated in the figure.

____________A

5)

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

__________V

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

1) at t = 0, the inductor acts as open circuit.

so, Req = R1 + R3 + R4

= 35 + 104 + 87

= 226 ohms

I4(0) = V/Req

= 12/226

= 0.0531 A

2) at t = infinite, the inductor acts as short circuit.

Req = R1 + R2*R3/(R2 + R3) + R4

= 35 + 35*104/(35 + 104) + 87

= 148 ohms

I4(inifnite) = V/Req

= 12/148

= 0.0811 A

3) IL(infinite) = I4(infinite)*R3/(R2 + R3)

= 0.0811*104/(35 + 104)

= 0.0607 A

4) Time constant of teh ckt, T = L/(R2+R3)

= 0.283/(35 + 104)

= 0.00204 s

= 2.04 ms

at time t

I3(open) = IL(infinite)*e^(-t/T)

= 0.0607*e^(-3.3/2.04)

= 0.01204 A

5) VL_max(closed) = I4(0)*R3

= 0.0531*104

= 5.52 V

6) VL_max(opened) = IL(infinite)*(R2 + R3)

= 0.0607*(35 + 104)

= 8.44 V

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