Nortons Theorem (ac)
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The methods described for Thevenin's theorem will each be
altered to permit their use with Norton's theorem. Since the Thevenin
and Norton impedances are the same for a particular network, certain
portions of the discussion will be quite similar to those encountered in
the previous section. We will first consider independent sources and the
approach developed in Chapter 8, followed by dependent sources and
the new techniques developed for Thevenin's theorem.
You will recall from Chapter 8 that Norton's theorem allows us to
replace any two-terminal linear bilateral ac network with an equivalent circuit consisting of a current source and an impedance, as in
Fig. 1.
The Norton equivalent circuit, like the Thevenin equivalent circuit, is
applicable at only one frequency since the reactances are frequency
dependent.
we find that
$$\bbox[10px,border:1px solid grey]{ I_{sc} = I_N} \tag{1}$$
and in Fig. 8(c) that
$$ E_{oc} = I_N Z_N$$
Or, rearranging, we have
$$ Z_N = {E_{oc} \over I_{sc}}$$
Fig. 1: The Norton equivalent circuit for ac networks.
Independent Sources
The procedure outlined below to find the Norton equivalent of a sinusoidal ac network is changed (from that in Chapter 8) in only one respect: the replacement of the term resistance with the term impedance.- Remove that portion of the network across which the Norton equivalent circuit is to be found.
- Mark the terminals of the remaining two-terminal network.
- Calculate $Z_N$ by first setting all voltage and current sources to zero (short circuit and open circuit, respectively) and then finding the resulting impedance between the two marked terminals.
- Calculate $I_N$ by first replacing the voltage and current sources and then finding the short-circuit current between the marked terminals.
- Draw the Norton equivalent circuit with the portion of the circuit previously removed replaced between the terminals of the Norton equivalent circuit.
Fig. 2: Conversion between the Thevenin and Norton equivalent circuits.
Example 1: Determine the Norton equivalent circuit for the network external to the 6-Ω resistor of Fig. 3.
Solution: Steps 1 and 2 (Fig. 4):
$$Z_1 = R_1 + j X_L = 3 Ω+ j 4Ω = 5Ω \angle 53.13^\circ$$
$$Z_2 = - j X_C = - j 5 Ω$$
Step 3 (Fig. 5):
$$ \begin{split}
Z_N &= {Z_1Z_2 \over Z_1 + Z_2}\\
&= {(5 Ω \angle 53.13^\circ)(5 Ω \angle -90^\circ) \over 3 Ω+ j 4Ω- j 5Ω} \\
&= {25 Ω \angle -36.87^\circ \over 3 -j 1}\\
&= {25 Ω \angle -36.87^\circ \over 3.16 \angle -18.43^\circ}\\
&= 7.91 Ω \angle -18.44^\circ = 7.50Ω - j 2.50 Ω\\
\end{split}
$$
Step 4 (Fig. 6):
Fig. 3: Example 1.
Fig. 4: Assigning the subscripted impedances to the
network of Fig. 3.
Fig. 5: Determining the Norton impedance for the
network of Fig. 3.
Fig. 6: Determining $I_N$ for the network of Fig. 3.
$$ I_N = I_1 = {E \over Z_1} = { 20 V \angle 0^\circ \over 5 Ω \angle 53.13^\circ }= 4 A \angle -53.13^\circ $$
Step 5: The Norton equivalent circuit is shown in Fig. 7.
Fig. 7: The Norton equivalent circuit for the network of Fig. 3.
Dependent Sources
As stated for Thevenins theorem, dependent sources in which the controlling variable is not determined by the network for which the Norton equivalent circuit is to be found do not alter the procedure outlined above. For dependent sources of the other kind, one of the following procedures must be applied. Both of these procedures can also be applied to networks with any combination of independent sources and dependent sources not controlled by the network under investigation. The Norton equivalent circuit appears in Fig. 8(a). In Fig. 8(b), Fig. 8: Defining an alternative approach for determining $Z_N$.
Example 2: Using one of the method described for dependent sources,
find the Norton equivalent circuit for the network of Fig. 9.
Solution:
For each method, $I_N$ is determined in the same manner. From Fig. 10,
using Kirchhoff's current law, we have
$$ 0 = I + hI + I_{sc}$$
or
$$ I_{sc} = -(1+h)I $$
Applying Kirchhoff's voltage law gives us
$$ E + IR_1 - I_{sc}R_2 = 0$$
and
$$IR_1 = I_{sc}R_2 - E$$
or
$$I = {I_{sc}R_2 - E \over R_1}$$
so
$$I_{sc} = -(1+h)I = -(1+h)({I_{sc}R_2 - E \over R_1})$$
or
$$R_1 I_{sc} = -(1+h)I = -(1+h)I_{sc}R_2 +(1+h)E $$
or
$$ I_{sc}[ R_1 +(1+h)R_2] = (1 + h)E$$
$$ I_{sc} = { (1+h)E \over R_1 + (1+h)R_2} = I_N$$
To find $Z_N$, we have to find with
$$Z_N = {E_{oc} \over I_{sc}}$$
$E_{oc}$ is determined from the network of Fig. 11.
By Kirchhoff's current law,
$$0 = I + hI \, \text{or}\, I(h + 1) = 0$$
For $h$, a positive constant $I$ must equal zero to satisfy the above.
Therefore, $$I = 0 \, \text{and}\, hI = 0$$ and $$ E_{oc} = E$$ with $$ \begin{split} Z_N &= {E_{oc} \over I_{sc}}\\ &= {E \over { (1+h)E \over R_1 + (1+h)R_2}}\\ &= {R_1 + (1+h)R_2 \over (1+h)}\\ \end{split}$$
Fig. 9: Example 2.
For each method, $I_N$ is determined in the same manner. From Fig. 10,
Fig. 10: Determining $I_{sc}$ for the network of Fig. 9.
Fig. 11: Determining $E_{oc}$ for the network of Fig. 9.
Therefore, $$I = 0 \, \text{and}\, hI = 0$$ and $$ E_{oc} = E$$ with $$ \begin{split} Z_N &= {E_{oc} \over I_{sc}}\\ &= {E \over { (1+h)E \over R_1 + (1+h)R_2}}\\ &= {R_1 + (1+h)R_2 \over (1+h)}\\ \end{split}$$
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