In the figure below, z = 150 V, R_1 = 10.0 ohm, R_2 = 22.0 ohm, R_3 = 30.0 ohm,

Question

In the figure below, z = 150 V, R_1 = 10.0 ohm, R_2 = 22.0 ohm, R_3 = 30.0 ohm, and L = 2.50 H. What are the values of the currents a very short time (i.e. nearly immediately) after switch S is closed? (a) i_1 A (b) i_2 A What are the values a long time late? (c) i_1 A (d) i_2 A The switch is then reopened, Just, then, what are the values of the currents? (e) i_1 A (f) _2 A What are the these values a long time late? (g) i_1 A (h) i_2 A A battery is connected to a series RL circuit at time t = 0. At what multiple of tau_L will the current be 0.540% less than its equilibrium value? Tau_L

Explanation / Answer

at t=0 inductor acts like a broken wire

And t = infinity inductor acts like a conducting wire. Considering these

Facts in mind.

Solve part a b c d

A) I1 = v/ r1+r2

= 150v/ (10+22) ohm. = 4.6875 A ans

B) I2 = I1 = 4.6875 A. ans

c) after long time

R eq = R1 +( R2x R3)/R2+R3

= 10 + 12.69 =22.69 ohm

I1 = v/R eq = 150/22.69 =6.69 A ans

d)I2 =I1 x R3/(R2+R3)

= 3.87 A ans

please post maximum 4 subparts in a question.i hope this was helpful.thanks.

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