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.
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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.
At maximum power dissipation,
RLOAD = RTH
So as it given in the problem, at maximum power dissipation Rload = 3kΩ
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.
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).
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
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