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}.
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 = I

^{2}R
81mW = I

^{2}X 9KΩ
I

^{2}= 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 R

_{1}now.
(R

_{1}//12kΩ) + 5kΩ = 9kΩ
R

_{1}//12KΩ = 4KΩ

__R___{1}= 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 V

_{1}. 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-V

_{1})/6kΩ + 6mA = 0
30/12kΩ + (30-V

_{1})/6kΩ + 6mA = 0
30/12kΩ + 30/6kΩ + 6mA = V

_{1}/6kΩ

__V___{1 }= 81V
If you have any problem please leave a comment.

Pabindu lakshitha

B.Sc (Engineering Undergraduate)

In fig 14.12 a how did you determine resistance in the middle branch and the total current?

ReplyDeleteIt could also be 4.8A and 15ohm.

Question says : The maximum power dissipated at Rload is measured as 81mW when the resistance R is set to 9kΩ....

DeleteAccording to maximum power transfer theorem, internal resistance = load (at maximum power)

So 14.12a shows that circuit.

NOTE: that is not related to 14.12. figue 14.12a shows the equivalent circuit for a circuit which dissipate maximum power for 9ohms load

its very thanks

ReplyDelete