مكافئ نورتن Norton Equivalent

مكافئ نورتن Norton Equivalent 4

مكافئ نورتن Norton Equivalent

يمكن استخدامه لتحديد مكافئ للدوائر الكهربية و لكن في صورة مصدر تيار مع مقاومة علي التوازي. ويمكن استنتاجه من مكافئ سفننس كالتالي :

I = Vth / Rth

و يوضح الشكل التالي التناظر بين مكافئ سفننس و مكافئ نورتن.
التناظر بين مكافئ سفننس و مكافئ نورتن
Example 1

Determine the current x in the 4-ohm resistance in the circuit shown in Fig. (8(a)) below.

Solution:

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Determine the current x in the 4-ohm resistance in the circuit shown  in Fig. (8(a)) below

Next, multiplying equation (ii) by 4 and equation (iii) by 5 and subtracting equation (iii) from equation (ii), we get

-28 x – 45 z = -80

or   28 x + 45 z = 80                                                          …(v)

Multiplying equation (iv) by 45 and equation (v) by 95 and adding the two, we get :      2840 x = 11650

or  284 x = 1165

x = 1165 / 284 = 4.1 A

 

Example2

Find I1, I2 and I3 in the network shown in Fig. (9), using loop-current method. Numbers against resistances indicate their values in ohms.
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Solution:

Different loops would be taken one after another.

Loop ABCDA:

– 10 I1 – 20(I1 – I2) –10 = 0

or         3 I1 –2I2 = – 1              …(i)

Loop BEFCB:

40 – 20 I2+ 10 – 10 (I2 – I3) – 20(I2 – I1) = 0

or         2 I1 –5I2 + I3 = – 5                   …(ii)

Loop EGHFE:

– 10 I3+50 –10(I3 – I2) – 10 = 0

or         I2– 2 I3= – 4                            …(iii)

1-               Multiplying equation (ii) by 2 and adding it to equation (iii), we get

4 I1 – 9 I2= – 14                       …(iv)

Solving for I1 and I2 from equation (i) and (iv), we get:

I1 = 1A            and      I2 = 2A

Solving these values in equation (iii), we have, I3=3A

 

Example 3

With reference to the network of Fig. (10), (where the number against resistances indicate their values in ohms and the internal resistance of the battery is given 1 Ω) by applying Thevenin’s theorem, find the following:

  1. (i) The equivalent e.m.f. of the network when viewed from terminals A and B.
  2. (ii) The equivalent resistance of the network when looked into from terminals A and B.
  3. (iii) Current in the load resistance RL of 15Ω

Solution:

  • Current in the network after load resistance has been removed [Fig. (10(b))]=24/(12+3+1)=1.5A

Then voltage across terminals AB=V=12*1.5=18V

Hence, so far as terminals A and B are concerned, the network has an e.m.f. of 18 volt (and not 24V).

(ii) There are two parallel paths between points A and B [Fig. (11(a))]. Imagine that battery of 24V is removed but not its internal resistance. Then equivalent resistance of the circuit as looked into from points A and B is

R=12*4/(12+4)=3 Ω

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(iii) When load resistance of 15Ω is connected across the terminals, then the network is reduced to the structure shown in Fig. (2.9(b)). Then  I=18/(15+3)=1A.
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Example 4

Three resistances R, 2R and 3R are connected in delta, Fig. (12(a)). Determine the resistances for an equivalent star connection. In Fig. (13), 160 volts are applied to the terminals AB. Determine (a) the resistance between the terminals A and B and (b) the current.
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Solution:

The three resistances are joined in delta in Fig. (12(a)).

We have in Fig. (12(b))

R1 =      R x 3R /( R + 2R + 3R) = R/2

R2 =      R x 2R /6 R = R/3

R3 =     2R x 3R / 6R= R

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R1 =           60 x 100 /(60+100+40)          = 30 Ω

R2 =           100 x 40 / 200                        = 20 Ω

R3 =           40 x 60 / 200                           = 12 Ω

Then the network of Fig. (14) is reduced to a simple structure of Fig. (15(a)).

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