Given the circuit shown in Figure 28-56 , theinductor has negligible internal re

Question

Given the circuit shown in Figure 28-56, theinductor has negligible internal resistance and the switchS has been open for a long time. The switch is thenclosed. (a) Find the current in the battery, the currentin the 100- resistor, and the current in the inductorimmediately after the switch is closed. (b) Find thecurrent in the battery, the current in the 100- resistor,and the current in the inductor a long time after the switch isclosed. After being closed for a long time the switch is nowopened. (c) Find the current in the battery, the currentin the 100- resistor, and the current in the inductorimmediately after the switch is opened. (d) Find thecurrent in the battery, the current in the 100- resistor,and the current in the inductor after the switch is opened for along time.

http://ebooks.bfwpub.com/physse6e/figures/28_56.gift

this is the link for the image

Explanation / Answer

I2 is the current going through the branch on the right, I3 is the current going through the branch on theleft,
(a)Right after the switch is closed I3=0 because it has nothad time to get there yet
E - I1R1 -I2R2 = 0 Kirchoff rule on the right mini loop But you got I1= I2 + I3 but I3=0 so I1 = I2
You get I1 = I2 = E/(R1+R2=10/110 A
(b) -Ldi/dt + I2R2=0 I2=0 because the inductor is completely charged I1=E/R1=10/10=1A I1=I2+I3=0+I3=1A
(c) I3=1A because its fully charged I1=0 because no current is going to go through the middlebranch I1=I2+I3 I2=I1-I3=0-1A I2=-I3=-0.1A
(d) There is no more current in the circuit everything isequal to 0 everywhere
So I

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