## 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 I

_{N}. 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

I

_{N}= 16/4

__I___{N }= 4A
Now we are about to find the Norton’s resistance R

_{N}. see figure 14.10
It is very easy.

R

_{N }= (6//3) + 4

__R___{N}= 6Ω
Now we can redraw the
circuit as shown in figure 14.11

I

_{1}+ I_{2}= 4A ------- (1)
As the two resistors are connected parallel, the voltage
across them should be equal, therefore,

6I

_{1 }= 12I_{2}
I

_{1}= 2I_{2 }----------- (2)
By (1) & (2)

__I___{2}= 4/3A
Q4) In order to find the V

_{1}& R_{1}values in the circuit shown in Figure14.12, power dissipation at the variable resistor R_{load}is measured, for the different values of R_{load}. The maximum power dissipated at R_{load }is measured as 81mW when the resistance R is set to 9kΩ. Find V_{1}& R_{1}.**See my next post for answers.**

**Please try to solve this problem at least three times.**

Hey how can u redraw the fig. 14.11 and where is the 4ohm resistance..

ReplyDeletecan u explain it please

i cant understand

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.

ReplyDeletethanx a lot

ReplyDelete