Day 9 Notes Norton and Max Power[1]

8/3/2019 Day 9 Notes Norton and Max Power[1]

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Nortons Theorm

Example:

Given the circuit shown below, obtain the Norton equivalent as viewed from terminals:

A-B

RN= 4 || (2 + 6 || 3) = 4 || 4 = 2

For IN consider the following:

After source transformations this becomes:


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Applying KVL to the circuit

-40 + 8i + 12 = 0 -> i = 3.5 A

HenceVth = 4i = 14V

And

IN = Vth/RN = 14 / 2 = 7 A

Now find RN and IN between point c and d.


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Maximum Power Transfer

In many practical situations, a circuit is designed to provide power to a load. While for

electric utilities, minimizing power losses in the process of transmission and distribution

is critical for efficiency and economic reasons, there are other applications in areas suchas communications where it is desirable to maximize the power delivered to a load. We


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now address the problem of delivering the maximum power to a load when given a

system with known internal losses. It should be noted that this will result in significant

internal losses greater than or equal to the power delivered to theload.

The Thevenin equivalent is useful in finding the maximum powera linear circuit can deliver to a load. We assume that we can

adjust the load resistance RL . If the entire circuit is replaced by its

Thevenin equivalent except for the load, as shown, the powerdelivered to the load is:

For a given circuit, V Th and RTh are fixed. By varying the load

resistance RL , the power delivered to the load varies as

sketched at right. We notice from that the power is small forsmall or large values of RL but maximum for some value of RL

between 0 and . We now want to show that this maximumpower occurs when RL is equal to RTh . This is known as themaximum power theorem.

Maximum power is transferred to the load when the load resistance equals the

Thevenin resistance as seen from the load (RL = RTh ).

To prove the maximum power transfer theorem, we differentiate p in the equation above

with respect to RL and set the result equal to zero. We obtain

This implies that

0 = (RTh + RL 2RL ) = (RTh RL )

which yieldsRL = RTh

showing that the maximum power transfer takes place when the load resistance RL equals

the Thevenin resistance RTh . We can readily confirm that this gives the maximum powerby showing that d2p / dRL

2 < 0

The maximum power transferred is obtained by substituting RL = RTh into the equation forpower, so that

This equation applies only when RL = RTh . When RL != RTh , we compute the power

delivered to the load using the original equation.


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Example:

(a) For the circuit given, obtain the Thevenin equivalent

at terminals a-b.(b) Calculate the current in RL = 8

(c) Find RL for maximum power

(d) Determine that maximum power.


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Find the value RL for maximum power transfer in the circuit shown below:

We need to find the Thevenin resistance R Th and the Thevenin voltage V Th across theterminals a-b. To get R Th , we source transformations and obtain:

To get V Th , we consider the circuit as shown.

Applying mesh analysis, 12 + 18i1 12i2 = 0, i2 = 2 A


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Solving for i1 , we get i1 = 2/3. Applying KVL around the outer loop to get V Th across

terminals a-b, we obtain

12 + 6i1 + 3i2 + 2(0) + V Th = 0 V Th = 22 V

For maximum power transfer,RL = RTh = 9 Ohms

and the maximum power is

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Day 9 Notes Norton and Max Power[1]