Chapter 30, Problem 054 In the figure, = 128 V, R1 = 14.5 , R2 = 19.0 , R3 = 27.

Question

Chapter 30, Problem 054 In the figure, = 128 V, R1 = 14.5 , R2 = 19.0 , R3 = 27.3 , and L = 2.85 H. Immediately after switch s is closed, what are (a) and (b) i27(Let currents in the indicated directions have positive values and currents in the opposite directions have negative values.) A long time later, what are (c) /i and (d) 2? The switch is then reopened. Just then, what are (e) and (f) i2? A long time later, what are (9) and (h) i2? Ri Rs RL (a) Number (b) Number (c) Number (d) Number (e) Number (f) Number Units Units Units Units Units Units (al Numher

Explanation / Answer

given that

E = 128 V

R1 = 14.5 ohm

R2 = 19 ohm

R3 = 27.3 ohm

L = 2.85 H

(a)

when the switch is closed , inductor will act as open , so no current will flow through branch of R3
so i1 = E/(R1 + R2) =128/(14.5 + 19) = 3.82 A

(b)

i2 is same as i1 .(because no current going through R3).

(d)

in this part, it is given that the switch is made closed long enough to overcome the transient response of inductor
it was made close enough to make it in steady state,

then voltage across R2 across the long time

VR2= E*[(R2 || R3)/{(R2 || R3) + R1}] = 128(11.20)/(11.20+14.5) = 55.65 V

so i2 = 55.65/(R2) = 55.65/19 = 2.92 A

(c)

in this case too inductor will act as short

so net resistance = R1 + (R2 || R3) ie = 14.5 +[(19*27.3)/(19+27.3)] = 25.7ohm

so i1= 128/25.7 = 4.98 A

(e)

after the reopening, there will be no current in R1 branch so i1 will be 0A.

(f)

after the reopening, there will be no current in R2 branch so i2 will be 0A.

(g)

as switch is still closed the value of i1 will be 0 A

(h)

after a long time inductor would get discharged and thus will again act as open circuit

so curent i2 = 0A

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