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 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 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 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 = V²/R     to  R_LOAD

147mW = V²/ 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 V₀

V = V₀ + V₁  (see figure 12.3)

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

Applying nodal equation to point B

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

-V₁/4 + (-42)/3 +13 = 0

-V₁/4 -1 = 0

V₁  = -4v -------------(2)

substitute (2) on (1)

V₀  = 24v

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pabindu lakshitha
B.Sc(engineering undergraduate)