#### SOLVED PROBLEMS (2)

Q4) In order to determine the unknown voltage V

_{0}and the resistance R in the circuit shown in figure 12.1 , power dissipation at the variable resistor R_{L}is measured, for the different values of R. the maximum power dissipation at the variable resistor R_{L}is measured as 147mW when the resistance R_{L}is set to 3kΩ. Determine the values of V_{0 }& R.

**It is reccomended to workout previous problems before this. And you should try to solve this problem without looking at the answers given below. Don’t even look at single word in the answers if you didn’t try to solve it atleast three times.**

Answer:

It says that the maximum power dissipation at R

_{L}occurs at 3kΩ. So you may understand that the thevenin resistance R^{TH}is also 3kΩ. This is one of the most important practical usages of thevenin’s theorem. This is the combination of both maximum power transfer theorem and thevenin’s theorem.
If you still can’t understand this method, consider the tehvenin
equivalent circuit with the load connected. Now this circuits it exactly equivalent
to the original circuit given. So instead of the given circuit we can use the
thevenin equivalent circuit with load connected. Now, you may see that there
are only two resistances available (R

_{LOAD }and R^{TH}). According to maximum power transfer theorem, to get the maximum power output these two resistance values should be equal.
This means,

At maximum power dissipation,

**R**

_{LOAD}= R^{TH}
So as it given in the problem, at maximum power dissipation R

_{load}= 3kΩ
Therefore

R

^{TH}= 3KΩ
Now we can find the resistance ‘R’

From figure 12.2

[ (R+2) // 3 ] + 1 = 3 (all values in kΩ)

[ (R+2)*3 / R+5 ] + 1 = 3

3R + 6 = 2R + 10

__R = 4Ω__
Now we have find R.

We have given that the maximum power dissipation in R

_{LOAD}is 147mW at R_{LOAD}= 3.
Apply,

p = V

^{2}/R to R_{LOAD}
147mW = V

^{2}/ 3kΩ

**V = 21v**

**This is the voltage across load when the maximum power dissipates, See figure 12.2(a).**

**So**

**V**

^{TH}/2 = 21V

__V__^{TH}= 42V
Now see figure 12.3

**Note that V**

_{C}= V^{TH}= 42V
We have grounded the point B.

Apply nodal equation to point C

(V

_{C}– V)/2kΩ + (V_{C}– 0)/3kΩ – 12mA – 13mA = 0
(42 – V) /2kΩ + 42/3kΩ
-25mA = 0

(42 – V) x 10

^{-3}/2 = 11x10^{-3}
42 – V = 22

**V = 20v**

But this is not what we need. We need to find V

_{0}
V = V

_{0}+ V_{1}(see figure 12.3)**V**

_{0}+ V_{1 }= 20 ---------------------------(1)

Applying nodal
equation to point B

(0 – V

_{1})/ 4kΩ + (0 - V_{C})/ 3kΩ + 13mA = 0
-V

_{1}/4 + (-42)/3 +13 = 0
-V

_{1}/4 -1 = 0**V**

_{1 }= -4v -------------(2)

substitute (2) on (1)

__V___{0 }= 24v

pabindu lakshitha

B.Sc(engineering undergraduate)

Thank you,,

ReplyDeleteThis comment has been removed by a blog administrator.

ReplyDeletei dnt understand why vth had to be divided by 2

DeleteYou should study about maximum power transfer theorem, then u will get to know about it.

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