Norton's theorem solved problems(3)

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.

Answer:

You have to solve this problem from the end to beginning. This means we must go backward in the problem.

As the problem says, the maximum power output occurs when R_LOAD = 9kΩ. Therefore according to the maximum power transfer theorem, internal resistance as seen from A-B terminals is also equal to 9kΩ. So this resistance is therefore equal to the Norton’s resistance of A-B terminals.

As the maximum power dissipation at R_LOAD = 9kΩ is equal to 81mW, we can find the current flowing through the load resistance R_LOAD .

P = I²R

81mW = I² X 9KΩ

I² = 9 X 10^-6

I = 3mA

So now we can draw the Norton’s equivalent circuit as shown is figure 14.12.a to determine the Norton’s current.

We can simply understand the Norton’s current is equal to 6mA, as the two resistance are equal and parallel connected.

Now we know the Norton’s current.

So now see figure 14.13.

As the Norton’s resistance is equal to 9kΩ, we can obtain the value of R₁ now.

(R₁//12kΩ) + 5kΩ = 9kΩ

R₁//12KΩ = 4KΩ

R₁ = 6KΩ

Now, we are about to draw the figure that we draw when we wants to find the Norton’s current. Of course we know the Norton’s value. But here we draw it to find V₁. See figure 14.14

So as we know the value of I_N, we can find the voltage across 5kΩ which is equal to the voltage at point c (V).

Ohm’s law to 5kΩ resistor,

V = IR

V = 6mA x 5kΩ

V = 30V

Now by applying nodal equation on point C,

V/12kΩ  + (V-V₁)/6kΩ + 6mA = 0

30/12kΩ + (30-V₁)/6kΩ + 6mA = 0

30/12kΩ + 30/6kΩ + 6mA = V₁/6kΩ

V₁ = 81V

If you have any problem please leave a comment.

Pabindu lakshitha

B.Sc (Engineering Undergraduate)