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.
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 RLOAD = 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 RLOAD = 9kΩ is equal to 81mW, we can find the current flowing through the load resistance RLOAD .
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 IN, 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.
B.Sc (Engineering Undergraduate)