As shown in the figure, = 100.0 V, R1 = 14.0 , R2 = 24.0 , R3 = 36.0 and L = 5.0
ID: 1969731 • Letter: A
Question
As shown in the figure, = 100.0 V, R1 = 14.0 , R2 = 24.0 , R3 = 36.0 and L = 5.0 H.
a. What is i1 immediately after switch S is closed? (Let currents in the indicated directions have positive values and currents in the opposite directions have negative values)
b. What is i2 immediately after switch S is closed?
c. A long time later, what is i1?
d. A long time later, what is i2?
e. The switch is then reopened.
Just then, what is i1?
f. Just then, what is i2?
g. And a long time later, what is i1
h. And a long time later, what is i2?
Explanation / Answer
a.Immediately after the switch is closed: the inductor t(0-) has no current through it. since current in an inductor cannot change instantaneously ,the branch of inductor acts like open circuit at t=t(0+) hence i1=E/(R1+R2)=100/(14+24)=2.632 Amps at t(0+) *note t(0) denotes switching t(0+) is just after switching and t(0-) is just before b.i2,from the above logic at t(0+) = i1 hence i2=i1=2.632 Amps c.A long time later the inductor comes to steady state(meaning avg voltage across L is zero) and acts like a short. hence i1=E/(R1+(R2||R3))=100/(14+(24||36))= 100/28.4=3.5211 Amps d.i2 is current through R2 = [R3/(R2+R3)]*i1=[36/60]*3.5211=2.1126 Amps e.When the switch is reopened there is no exit path (meaning only one node is connected to a network) for the current i1 hence the current is zero amps.(note u need to nodes to define a path for current) f.The logic is same as in a) current through inductor cannot change instantaneously so current through just after switch is reopened is i3= 3.5211-2.1126=1.4085 Amps the same current should flow through R2 to complete the current flow path. hence i2=-i3=-1.4085 Amps note the - sign indicates the current flows in reverse direction to as assumed in figure g.After a long time later again the inductor acts like a short and there is voltage source exciting R2, R3 so current is zero again i1=i2=zero amps
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