Method I:

Norton Equivalent current (I_SC) is the short circuit current across the load.

The circuit to evaluate the Norton current is drawn as:

From the given curve, when V = 0, I = - 6 A

Since I_SC = - I

I_SC = 6A

Since the Norton and Thevenin equivalents are dual of each other, we can write:

\({{R}_{th}}=\frac{{{V}_{th}}}{{{I}_{sc}}}\)

Where V_th is the open-circuit voltage at the load

From the given characteristic curve, when I = 0 (Open-circuit), V_th is:

V_OC = V_th = 3 V

∴ The Norton equivalent resistance will be:

\({{R}_{N}}={{R}_{th}}=\frac{3}{6}=0.5~\text{ }\!\!\Omega\!\!\text{ }\)

Method II:

From the given graph, the equation of the line is:

I = 2V – 6     ...1)

For the given linear network across the load, the Norton equivalent circuit can be drawn as:

From the above figure, we can write:

\({{I}_{N}}+I=\frac{V}{{{R}_{N}}}\)

V = Voltage across the load

\(I=\frac{V}{{{R}_{N}}}-{{I}_{N}}\)

Comparing this with equation (1), we can write:

\(\frac{1}{{{R}_{N}}}=2\)

R_N = 0.5 Ω

I_N = 6A

According to Thevenin’s theorem, any linear circuit across a load can be replaced by an equivalent circuit consisting of a voltage source V_th in service with a resistor R_th as shown:

V_th = Open circuit Voltage at a – b (by removing the load), i.e.

If a linear circuit contains dependent sources only, i.e., there is no independent source present in the network, then the open-circuit voltage or Thevenin voltage will simply be zero. (Since there is no excitation present)

Important Note:

To evaluate the Thevenin Resistance, we consider the following two cases:

Case 1: Circuits with independent Sources only

If the network has no dependent sources, we turn off all independent sources. R_th is the input resistance of the network looking between terminals a and b.

Case 2: Circuit with Both Dependent and independent sources

We may use one of the two methods:

Using a Test Source:

We apply a voltage source V₀ at terminals a and b and determine the resulting current I₀.

Then R_th = V₀/I₀

Using Short circuit current:

• Connect a short-circuit between terminal a and b at the load.
• Do not set independent sources zero in this method because we have to find short-circuit current.
• Now obtain the short-circuit current I_SC through terminals a, b.
• The Thevenin resistance will then be given by R_Th = V_Th/I_SC.

Taking KVL in the loop

10 – 2i – 5 – 8i = 0

i = 0.5 A

⇒ V_th = 10 – 8 (0.5) = 6 V

To find R_Th, shorting the voltage sources

R_Th = 8 // 2 Ω = 1.6 Ω

Reciprocity Theorem:

In any passive linear bilateral network, if a single voltage source ‘V’ in the branch AB produces the current response ‘I’ in the branch CD, then the removal of voltage source from the branch AB and its insertion in the branch CD will produce same current ‘I’ in the branch AB.

In the reciprocity theorem, if the position of excitation and responses are interchanged, then their ratio remains the same.

\(\frac{{{V_1}}}{{{I_1}}} = \frac{{{V_2}}}{{{I_2}}}\)

Application:

In the reciprocity theorem, the polarity of the voltage source should have the same correspondence with the branch current in each of the circuits.

The polarity of the voltage source and current direction is as shown:

So, \(\frac{{{V}_{1}}}{{{I}_{1}}}=\frac{-{{V}_{2}}}{{{I}_{2}}}=\frac{{{V}_{3}}}{{{I}_{3}}}\)

\(\Rightarrow \frac{10}{2.5}=\frac{-20}{{{I}_{2}}}=\frac{40}{{{I}_{3}}}\)

Here, we cannot use superposition theorem to calculate the net power as power is a non-linear quantity

Since P = I²R

\(I=\pm \sqrt{\frac{P}{R}}\)

When only E₁ is acting alone, the power consumed by the resistance R is 18 W, i.e.

\(18=I_{1}^{2}R\)

\({{I}_{1}}=\pm \sqrt{\frac{18}{R}}\)

Similarly, we can write

\({{I}_{2}}=\pm \sqrt{\frac{50}{R}}\)

\({{I}_{3}}=\pm \sqrt{\frac{98}{R}}\)

When all the sources are acting simultaneously the net current through the resistance R using Superposition theorem is:

I = ± I₁ ± I₂ ± I₃

\(I=\pm \sqrt{\frac{18}{R}}\pm \sqrt{\frac{50}{R}}\pm \sqrt{\frac{98}{R}}\)

Net power when all the sources are acting simultaneously will be:

P_net = I²R

\({{P}_{net}}={{\left( \pm \sqrt{\frac{18}{R}}\pm \sqrt{\frac{50}{R}}\pm \sqrt{\frac{98}{R}} \right)}^{2}}R\)

\({{P}_{net}}={{\left( \pm \sqrt{18}\pm \sqrt{50}\pm \sqrt{98} \right)}^{2}}\) 

\({{P}_{net}}={{\left( \pm 3\sqrt{2}\pm 5\sqrt{2}\pm 7\sqrt{2} \right)}^{2}}\)

The maximum value of the power will happen when all the current will be contributing in the same direction:

i.e. \({{P}_{max}}={{\left( 3\sqrt{2}+5\sqrt{2}+7\sqrt{2} \right)}^{2}}\)

\({{P}_{max}}={{\left( 15\sqrt{2} \right)}^{2}}\)

P_max = 450 W

For minimum power consumed by R, the current contribution by I₁ and I₂ must be same, but opposite to the current contribution by I₃, i.e.

\({{P}_{min}}={{\left( +3\sqrt{2}+5\sqrt{2}-7\sqrt{2} \right)}^{2}}\)

\(or~{{P}_{min}}={{\left( -3\sqrt{2}-5\sqrt{2}+7\sqrt{2} \right)}^{2}}\)

\({{P}_{min}}={{\left( \sqrt{2} \right)}^{2}}\)

Concept:

The maximum power theorem for DC circuit states that, for a variable load, maximum power is transferred to it when:

RL = Rth

Rth = Thevenin's Equivalent resistance across the load.

Application:

Since we have a dependent source, we place a test current source I₀ as shown:

From the above figure, we can write:

V_x = -2I₀

Also, Applying KVL through the loop, we get:

kV_x - V_x + 4I₀ – V₀ + 4I₀ = 0

⇒ V_x (k - 1) + 8I₀ = V₀

⇒ -2I₀ (k - 1) + 8I₀ = V₀

\(\frac{{{V}_{0}}}{{{I}_{0}}}={{R}_{th}}=8-2~\left( k-1 \right)\)

Given R_L = 2Ω

For maximum power to be transferred to R_L,

R_L = R_th, i.e.

2 = 10 – 2k

Thevenin voltage is the open-circuit voltage at the load.

Since the load is open-circuited, V_x = V_th

Applying KCL at node V_x, we get:

\(\Rightarrow \frac{{{V}_{x}}-2{{V}_{x}}}{3}+\frac{{{V}_{x}}}{6}+1=0\)

⇒ -2V_x + V_x = -6

V_x = 6 V

∴ V_th = 6V

Now, to evaluate the Thevenin’s Resistance, there are two methods that can be used:

Method 1 (Using a Current source at the load):

Since we have a dependent source present in the circuit, we put a test current source I₀ as shown:

Applying KCL at node V_x we get

\(\Rightarrow \frac{{{V}_{x}}-2{{V}_{x}}}{3}+\frac{{{V}_{x}}}{6}-{{I}_{0}}=0\)

\(\Rightarrow \frac{-{{V}_{x}}}{6}={{I}_{0}}\)

V_x = - 6I₀     ...1)

Applying KVL from V_x node to V₀, we get:

V_x + 3I₀ = V₀

Using equation (1),

-6I₀ + 3I₀ = V₀,

\(\frac{{{V}_{0}}}{{{I}_{0}}}={{R}_{th}}=-3\text{ }\!\!\Omega\!\!\text{ }\) 

Method II (Short cut):

Since the Thevenin and Norton equivalent circuits are related by Superposition Theorem, we can write:

\({{R}_{TH}}=\frac{Open~circuit~voltage}{Short~circuit~current}\)

\({{R}_{th}}=\frac{{{V}_{th}}}{{{I}_{SC}}}\)

To obtain the Thevenin Resistance, we first find the short circuit current through the load as shown:

Applying KCL at node V_x, we get

\(\Rightarrow \frac{{{V}_{x}}-2{{V}_{x}}}{3}+\frac{{{V}_{x}}}{6}+1+\frac{{{V}_{x}}}{3}=0\)

\(\Rightarrow \frac{{{V}_{x}}}{6}=-1\)

V_x = -6V

\(\therefore ~{{I}_{SC}}=\frac{{{V}_{x}}}{3}=\frac{-6}{3}=-2A\) 

When V₂ = 0 V

I₂ = -6 A

By current division,

\({I_2} = - {I_1}\left( {\frac{6}{{6 + 2}}} \right)\)

\(\Rightarrow - 6 = - {I_1}\left( {\frac{6}{8}} \right) \Rightarrow {I_1} = 8\;A\)

By applying KCL at node ‘O’,

I₃ + I₂ + I₁ = 0

⇒ I₃ = -(I₁ + I₂) = -(8 - 6) = -2 A

When V₁ = 0 V

I₁ = 6 A

By current division,

\({I_1} = - {I_2}\left( {\frac{6}{{3 + 6}}} \right)\)

\(\Rightarrow 6 = - {I_2}\left( {\frac{6}{9}} \right) \Rightarrow {I_2} = - 9\;A\)

By applying KCL at node ‘O’,

I₁ + I₂ + I₃ = 0

⇒ I₃ = -(I₁ + I₂) = -(6 - 9) = 3 A

When both sources are applied,

I₃ = -(I₁ + I₂) = -(6 - 9) = 3 A

When both sources are applied,

The given circuit is drawn as shown:

Thevenin Analysis:

Thevenin Voltage is simply the open-circuit voltage across the load.

Since the voltage across the load terminal is V_s and it is independent of any other current or voltage, the Thevenin equivalent voltage will be:

V_th = V_s

In calculating the Thevenin Resistance we short circuit the voltage source and open circuit the current source.

Since the given voltage source is ideal, the internal resistance will be 0. The circuit is then redrawn as:

R_th = 0

So, the Thevenin Equivalent circuit will exist with V_th = V_s and R_th = 0 as:

So the Thevenin equivalent circuit exists and it is simply that of a voltage source.

Norton Analysis:

In calculating the Norton current we short circuit the load as shown:

V_s is connected parallel to a short circuit load. The Norton current I_sc will be:

\(I_{sc}=\frac{V_s}{0}=\infty\), which is not possible.

Also, the Norton Resistance is the same as the Thevenin Resistance and is 0 Ω

The Norton Equivalent circuit will be:

Since the parallel resistance R is 0 Ω, all the current will pass through it making the load current I_L= 0A. This is again not possible because clearly in the given circuit we have a voltage source in parallel to the load that will result in the flow of some current through it.

We, therefore, conclude that the Norton equivalent analysis is not valid in this case. 

To maximize the power transfer through the load, IL should be maximum.

Let us find expression for I_L by using superposition theorem.

When only 10 V source is active.

\(I_L' = \frac{{10}}{{R + 150}}A\)

When 10 A source is active   

By using current division:

\(I_L^{''} = \frac{{10\;R}}{{R + 150}}A\)

Now, the current flows through R_L,

\({I_L} = \frac{{10}}{{R + 150}} + \frac{{10\;R}}{{R + 150}}\)

\(I_L=\frac{10+10R}{R+150}\)

\(I_L=\frac {10+10/R}{1+150/R}\)

For R = 0, I_L will be:

\(I_L=\frac{10}{150}A\)

For large values of R, I_L approaches 10 A.

So, the maximum value of I_L occurs at a maximum value of R.

Hence, R should be as large as possible, i.e.

⇒ R = infinite

Common Mistake:

Maximum power theorem states that for maximum power to be transferred to the load resistance RL, RL must equal the Thevenin Equivalent resistance, i.e.

RL = Rth

But here, we are asked to find the value of R, and not RL that will result in the maximum power to be transferred to load RL. So we cannot go by the standard procedure of equating RL with the Thevenin equivalent resistance.