problem 2 on superposition

Thread Starter

electronicstech07

Joined Nov 16, 2007
18
I have attached my work on another problem involving superposition. It involves a Is and Vs and is asking me to find Ir3. Help is much appreciated!
 

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vvkannan

Joined Aug 9, 2008
138
While calculating I (killing Vs) entering at B the resistance in the denominator should be the sum of total resistance.I think you have made a mistake there.
 

Thread Starter

electronicstech07

Joined Nov 16, 2007
18
While calculating I (killing Vs) entering at B the resistance in the denominator should be the sum of total resistance.I think you have made a mistake there.
You're correct again. I guess I'm going a little too fast calculating the answer on some of these problems. Thanks again!
 

Thread Starter

electronicstech07

Joined Nov 16, 2007
18
What is the answer given in the book?

hgmjr
The answer is 1.6 mA and that is what I got when correcting the value of the denominator to 852.5 ohms when calculating for I entering pt. B.

I entering pt. B = 680/852.5 * 100 mA = 79.76 mA.

Using I divider, Ir3 = 220/1020 ohms * 79.76 mA = 17.2 mA

Solving for Ir3, 17.2-15.6 = 1.60 mA.
 

hgmjr

Joined Jan 28, 2005
9,029
Here is the Millman's Theorem approach to the problem.
Below is the Millman's Theorem expression for the voltage across the 680 ohm resistor as well as the series combination of the 330 and the 470 ohm resistor.

\(\LARGE V_{\large x}\,=\,\frac{\frac{-20}{220}\,+\,0.1}{\frac{1}{220}\,+\,\frac{1}{680}\ +\,(\frac{1}{330\,+\,470})}\)

To obtain the current in R3, all that is needed it to divide Vx by 800

\({\Large I}_{R3}\,=\,\frac{V_X}{800}\)
 
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