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 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

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 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 V₁ & R₁.

See my next post for answers.

Please try to solve this problem at least three times.

pabindu lakshitha

B.Sc(Engineering Undergraduate)