Sunday, September 2, 2012

THEVENIN'S THEOREM SOLVED PROBLEMS (2)


SOLVED PROBLEMS (2)




Q4) In order to determine the unknown voltage V0 and the resistance R in the circuit shown in figure 12.1 , power dissipation at the variable resistor RL is measured, for the different values of R. the maximum power dissipation at the variable resistor RL is measured as 147mW when the resistance RL is set to 3kΩ. Determine the values of V0 & 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 RL occurs at 3kΩ. So you may understand that the thevenin resistance RTH 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 (RLOAD and RTH). 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,

RLOAD = RTH

So as it given in the problem, at maximum power dissipation Rload = 3kΩ

Therefore
RTH = 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 RLOAD is 147mW at RLOAD = 3.

Apply,

 p = V2/R     to  RLOAD

147mW = V2/ 3kΩ

V = 21v

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



So

VTH/2 = 21V

VTH = 42V

Now see figure 12.3



Note that VC = VTH = 42V

We have grounded the point B.

Apply nodal equation to point C

(VC – V)/2kΩ  +  (VC – 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 V0

V = V0 + V1  (see figure 12.3)

V0 + V1 = 20 ---------------------------(1)

Applying nodal equation to point B

(0 – V1)/ 4kΩ  + (0 - VC)/ 3kΩ  + 13mA = 0

-V1/4 + (-42)/3 +13 = 0
-V1/4 -1 = 0
V1  = -4v -------------(2)

substitute (2) on (1)

V0  = 24v


 LEAVE YOUR PROBLEM AS A COMMENT.



pabindu lakshitha
B.Sc(engineering undergraduate)

4 comments:

  1. This comment has been removed by a blog administrator.

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    1. i dnt understand why vth had to be divided by 2

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

      Delete

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