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Saturday, September 8, 2012

Norton's theorem solved problems(2)


solved problems on Norton's theorem (2)


Q3) Using Norton’s theorem calculates the current flowing through 12Ω resistor. You may refer figure 14.8



Hope you tried individually to solve this problem as I mentioned in my previous post. I’ll show you how to solve this problem here.

First we remove the 12Ω resistor and short circuit the two terminals to find the Norton’s current as shown in figure 14.9




There are many simple ways to find IN. I’ll use the nodal equation method. Therefore I grounded the point D.

Applying nodal equation to point D,

V/6 + V/4 + (V-36)/3 = 0

V/6 + V/4 + V/3 = 36/3

3V/4 = 12

V = 16V

Now apply Ohm’s law to 4Ω resistor,

I = V/R

IN = 16/4

IN = 4A

Now we are about to find the Norton’s resistance RN . see figure  14.10



It is very easy.

RN = (6//3) + 4

RN = 6Ω

 Now we can redraw the circuit as shown in figure 14.11




I1 + I2 = 4A ------- (1)

As the two resistors are connected parallel, the voltage across them should be equal, therefore,

6I1 = 12I2

I1 = 2I2 ----------- (2)

By (1) & (2)

I2 = 4/3A

Q4) In order to find the V1 & R1 values in the circuit shown in Figure14.12, power dissipation at the variable resistor Rload is measured, for the different values of Rload. The maximum power dissipated at Rload is measured as 81mW when the resistance R is set to 9kΩ. Find V1 & R1.




See my next post for answers.
Please try to solve this problem at least three times.



pabindu lakshitha
B.Sc(Engineering Undergraduate) 



3 comments:

  1. Hey how can u redraw the fig. 14.11 and where is the 4ohm resistance..
    can u explain it please
    i cant understand

    ReplyDelete
  2. we remove all the resistors in the circuit and replace by Norton resistance in parallel with the norton current source. 12ohm resistor is the LOAD resistance.

    ReplyDelete