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Taking the Measure   


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Try your hand at EE 101
February 27, 2006

In a book review in our March issue, I chide an author of an otherwise fine book who, in a chapter covering Thevenin's theorem, fails to provide the Thevenin equivalent of a sample circuit he presents. As I wish to be constructive and not just critical, I'll provide the equivalent circuit and it's derivation.

The circuit (click here to see the figure) consists of three parts:

First, a voltage source V1 with its low side grounded connects to the output terminal through series resistor R1.

Second, a voltage source V2 with its low side grounded connects to the output terminal thorugh series resistor R2.

Finally, a resistor R3 shunts the output terminal to ground.

I'll post the algebraic formulas for open-circuit output voltage VT and equivalent series impedance RT, with respect to R1, R2, R3, V1, and V2, on Monday March 6, 2006. Meanwhile, try your hand at one of these numeric cases:

1. R1=R2=R3=30 kilohms, V1=10 V, and V2=5 V.

2. R1=R2=100 kilohms, R3 = infinity, V1=10 V, and V2=5 V.

3. R1=R2=100 kilohms, R3 = infinity, V1=10 V, and V2=0 V.

Post your answer in the blog comment area.

Bonus question: what is the Norton equivalent of your submission?

Addendum: Find the answers here and more discussion here.


Posted by Rick Nelson on February 27, 2006 | Comments (10)


March 2, 2006
In response to: Try your hand at EE 101
Clyde commented:

Use Millmans's theorem for nonideal voltage sources. E=(E1/Z1 + E2/Z2)/(1/Z1 + 1/Z2), Z= 1/(1/Z1 + 1/Z2) (parallel combination). Electronics Engineers' Handbook, First Edition Copyright 1975, McGraw-Hill, Donald G. Fink, Editor-in-Chief, Section 3-69, P. 3-59.




March 2, 2006
In response to: Try your hand at EE 101
halvas commented:

To me it looks straightforward: Vt=V1*R2//R3/(R1+R2//R3)+ V2*R1//R3/(R2+R1//R3), Rt=R1//R2//R3. Using example 1, you get Vt=5V, Rt=10k. Using example 2, you get Vt=7.5V, Rt=50k. Using example 3, you get Vt=5V, Rt=50k. In this last example V2= 0V is virtually just a short, so you are calculating a simple voltage divider. As I see the result as obvious, I have tried ex1 in PSPICE, just to ensure there are no tricks involved. I am a lecturer in electronics, and I know the worst thing we can do to our students at an examination is to put a simple question with a simple answer--like this--because the students will often waste lots of time to find the "correct, difficult solution"--it can't be so obviously easy, there must be something hidden, they think.




March 2, 2006
In response to: Try your hand at EE 101
EE 101 Thevenin EQ Test commented:

1) i1=i3=1/6 A, i2=0 A, vt=v2=5 V. Thevenin EQ: v = 5V, R = 30 Ohms, i = 1/6A. 2) v = 5V, i =1/40A, R = 200 Ohms. 3) v = 10V, i = 1/20A, R = 200 Ohms.




March 3, 2006
In response to: Try your hand at EE 101
Snooker Dave commented:

The Thevenin circuit consists of a voltage source in series with a resistor. The Resistor is = to the parallel combination of all three resistors (R1||R2||R3). The Voltage source is the sum of two voltages (using superposition): V1 * R3||R2/(R1 + R3||R2) + V2 * R3||R1/(R2 + R3||R1). Send my prize to: daverutkowski@yahoo.com.




March 3, 2006
In response to: Try your hand at EE 101
Serge commented:

The solution does not really depend on the resistor value--the fact that the resistors are equal (or that one is +Inf. in cases 2 and 3) is enough. Vt=(V1R2+V2R1)/(R1+R2+(R1R2/R3)). When R1=R2=R3, Vt = (V1+V2)/3; when R1=R2, and R3=+Inf, the formula is Vt= (V1+V2)/2. For the Norton question, the Ieq=V1/R1+V2/R2, and the Req=R1||R2||R3...




March 3, 2006
In response to: Try your hand at EE 101
Patrick O'Brien commented:

Vt=(V1/R1 + V2/R2)*(R1||R2||R3). This yields for 1: Vt=5v, Rt=10K. 2: Vt=7.5v, Rt=50K. 3: Vt=5v, Rt=50K.




March 4, 2006
In response to: Try your hand at EE 101
Dave Cooper commented:

The answer to this was so easy it took me all of about 3 minutes to get the solution. Vth = (R1R3V2 + R2R3V1)/[R1R2 + R1R3 + R2R3], Rth = R1||R2||R3. To get the Norton Equivalent, just do a source transformation.





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