Network Theorems

• Superposition Theorem:

  A generic system with input (stimulus, cause) $u$ and output (response, effect) $v$ can be described mathematically as a function $v=F(u)$. Superposition principle applies if this function is linear:
  

  $$F(ax+by)=aF(x)+bF(y)$$
  
  
  Superposition principle has two aspects:
  
  $$F(ax)=aF(x),\;\;\;\;\;\;F(x+y)=F(x)+F(y)$$
  
  
  The interpretation is that when the effect is directly proportional to the cause in a system $F$, then we can consider several causes ($x$ and $y$) individually and then combine the individual effects ($F(x)$ and $F(y)$). An electrical system of linear components (with linear voltage-current relation) is a linear system (if only voltage and current are of interest).

  When there exist multiple energy sources, the currents and voltages in the circuit can be found as the algebraic sum of the corresponding values obtained by assuming only one source at a time, with all other sources turned off (voltage sources treated as short circuit, current sources treated as open circuit).

  As superposition principle only applies to linear functions, it cannot be applied to nonlinear functions such as power (e.g., $V^2/R$ or $I^2R$).

  

  Superposition of voltage sources:
  

  $$I_{12}=\frac{aV_1+bV_2}{R}=a\frac{V_1}{R}+b\frac{V_2}{R}=aI_1+bI_2$$
  
  
  where $I_1=V_1/R$ ($V_2$ short circuit) and $I_2=V_2/R$ ($V_1$ short circuit). However, note that superposition principle does not apply to power:
  
  $$P_{12}=\frac{(V_1+V_2)^2}{R}\ne\frac{V_1^2}{R}+\frac{V_2^2}{R}=P_1+P_2$$
  
  
  Superposition of current sources:
  
  $$V=R(aI_1+bI_2)=aI_1R+bI_2R=aV_1+bV_2$$
  
  
  where $V_1=I_1R$ ($I_2$ open circuit) and $V_2=I_2R$ ($I_1$ open circuit). Again, superposition principle does not apply to power:
  
  $$P_{12}=R(I_1+I_2)^2 \ne RI_1^2+RI_2^2=P_1+P_2$$
  
  

  Example 1: The previous example can also be solved by superposition theorem.

  

  First turn the voltage source of 20V off (short-circuit with 0V), and get
  

  $$I'_1=\frac{32}{2+8 \vert\vert 4}=\frac{48}{7},\;\;\;I'_2=-\frac{32}{7}, \;\;\;I'_3=\frac{16}{7}$$
  
  
  Second turn the voltage source of 32V off and get
  
  $$I''_2=\frac{20}{4+8 \vert\vert 2}=\frac{25}{7},\;\;\;I''_1=-\frac{20}{7}, \;\;\;I''_3=\frac{5}{7}$$
  
  
  The overall currents can then be found as the algebraic sums of the corresponding values obtained with one voltage source turned on at a time:
  
  $$I_1=I'_1+I''_1=\frac{48}{7}-\frac{20}{7}=4,\;\;\;\; I_2=I''_... ...{32}{7}=-1,\;\;\;\; I_3=I'_3+I''_3=\frac{16}{7}+\frac{5}{7}=3$$
  
  

  Example 2: Find voltage $V$ and current $I$.

  

  First, we solve this problem using node-voltage method. Assume the currents $I_1$ (left branch), $I_0$ and $I_2$ (right branch) all leave the top node, where the voltage is $V$ (with respect to the bottom treated as ground). By KCL, we have
  

  $$I_1+I_2=\frac{V-V_0}{R_1}+\frac{V}{R_2}=-I_0,\;\;\;\;\mbox{i.e.,}\;\;\;\; R_2(V-V_0)+R_1V=-I_0R_1R_2$$
  
  
  Solving for $V$, we get:
  
  $$V=\frac{R_2V_0-I_0R_1R_2}{R_1+R_2} =V_0\frac{R_2}{R_1+R_2}-I_0\frac{R_1R_2}{R_1+R_2}$$
  
  
  and
  
  $$I=I_2=\frac{V}{R_2}=\frac{V_0}{R_1+R_2}-I_0\frac{R_1}{R_1+R_2}$$
  
  
  Next, using superposition theorem, we get
  • Find $V'$ and $I'$ with the current source off (open circuit with zero current):
    
    $$I'=\frac{V_0}{R_1+R_2},\;\;\;\;V'=I'R_2=V_0\frac{R_2}{R_1+R_2}$$
    
    

  • Find $V''$ and $I''$ with the voltage source off (short circuit with zero voltage):
    
    $$I''=-I_0\frac{R_1}{R_1+R_2},\;\;\;\;V''=-I''R_2=-I_0\frac{R_1R_2}{R_1+R_2}$$
    
    
    Both $I''$ and $V''$ have a negative sign as their direction and polarity are opposite to those of the assumed current and voltage.
  • Find the sum of the two:
    
    $$I=I'+I''=\frac{V_0-I_0R_1}{R_1+R_2},\;\;\;\; V=V'+V''=R_2I=(V_0-I_0R_1)\frac{R_2}{R_1+R_2}$$
    
    

• Thevenin's Theorem:

  In principle, all currents and voltages of an arbitrary network of linear components and voltage/current sources can be found by any of the three methods discussed previous, namely, the branch current method, the loop current method and the node voltage method.

  However, if only the current and/or voltage associated with one component are of interest, it is unnecessary to find voltages and currents elsewhere in the circuit. The methods considered below can be used in such situations.

  Any one-port (two-terminal) network of resistance elements and energy sources is equivalent to an ideal voltage source $V_T$ in series with a resistor $R_T$, where

  • $V_T$ is the open-circuit voltage of the network, and
  • $R_T$ is the equivalent resistance when all energy sources are turned off (short-circuit for voltage sources, open-circuit for current sources).

  

  If we are only interested in finding the voltage $V$ across and current $I$ through one particular resistor in a complex circuit containing a large number of resistors, voltage and current sources, we can ``pull'' the resistor out and treat the rest of the circuit as a Thevenin voltage source $(V_T, R_T)$, and use Thevenin's theorem to find $V$ and $I$.

  Proof:

  

  Assume with the load the network's terminal voltage and current are $V$ and $I$ respectively.

  • Replace the load by an ideal current source $I$ while keeping the terminal voltage $V$ the same (b), the voltage or current anywhere in the system should not be affected.

  • Find the terminal voltage $V$ in terms of the internal energy sources inside the network and the external current source by superposition principle:
    • When the external current source is off (open circuit), the terminal voltage $V'=V_{oc}$ is due only to the sources internal to the network (c).
    • When all internal sources are turned off (short-circuit for voltage sources, open-circuit for current sources), the terminal voltage $V''=-R_0I$ where $R_0$ is the equivalent resistance of the network with all energy sources off (d).
    The overall terminal voltage is $V=V'+V''=V_{oc}-R_0I$.
  As far as the port voltage $V$ and current $I$ are concerned, a one-port network is equivalent to an ideal voltage source $V_T=V{oc}$ equal to the open-circuit voltage across the port, in series with an internal resistance $R_T=R_0$, which can be obtained as the ratio of the open-circuit voltage and the short-circuit current.

• Norton's Theorem:

  Any one-port (two-terminal) network of resistance elements and energy sources is equivalent to an ideal current source $I_N$ in parallel with a resistor $R_N$, where

  • $I_N$ is the short-circuit current of the network, and
  • $R_N$ is the equivalent resistance when all energy sources are turned off (short-circuit for voltage sources, open-circuit for current sources).

  Proof: The proof of this theorem is in parallel with the proof of the Thevenin's theorem. Again, assume with the load the network's terminal voltage and current are $V$ and $I$ respectively.

  • Replace the load by an ideal voltage source $V$ while keeping the terminal current $I$ the same, the voltage or current anywhere in the system should not be affected.

  • Find the terminal current $I$ in terms of the internal energy sources inside the network and the external voltage source by superposition principle:
    • When the external voltage source is off (short circuit), the terminal current $I'=I_{sc}$ is due only to the sources internal to the network.
    • When all internal sources are turned off (short-circuit for voltage sources, open-circuit for current sources), the terminal current $I''=-V/R_0I$ where $R_0$ is the equivalent resistance of the network with all energy sources off.
    The overall terminal current is $I=I'+I''=I_{sc}-V/R_0$.
  As far as the port voltage $V$ and current $I$ are concerned, a one-port network is equivalent to an ideal current source $I_T=I{sc}$ equal to the short-circuit current through the port, in parallel with an internal resistance $R_N=R_0$, which can be obtained as the ratio of the open-circuit voltage and the short-circuit current.

  Of course, the Norton's theorem can also be easily proven by converting Thevenin's equivalent circuit of an ideal voltage source $V_T=V{oc}$ in series with a resistance $R_T$ to an equivalent circuit of an ideal current source $I_N=I_{sc}=V_{os}/R_T$ in parallel with a resistance $R_N=R_T$.

  Load Line and Output Resistance

  Due to the Thevenin's and Norton's theorems, any one-port network of resistors and energy sources can be converted into a simple voltage or current source with an internal or output resistance $R_0$. Moreover, the relationship between the voltage $V$ across and the current $I$ through the load is a straight line referred to as the load line. The slope $-\triangle V/\triangle I=R_0$ of the load line indicates the internal or output resistance of the network, as shown in the figure below. One the other hand, the input resistance $R_L$ of the load can also be represented on the graph as a straight line with its slope $\triangle V/\triangle I=R_L$. The intersection of these two lines indicates the actual voltage $V$ and current $I$ with the load $R_L$.

  

  The output resistance $R_0$ of a network can also be determined experimentally by varying the load $R_L$. Assume $(V_1, I_1)$ are associated with load $R_1$ and $(V_2, I_2)$ with load $R_2$, then the output resistance of the source network can be found to be:

  
  

  $$R_0=-\frac{\triangle V}{\triangle I}=-\frac{V_1-V_2}{I_1-I_2}$$
  
  

  To show this, we assume the source network is converted to a voltage source with $V_0$ and $R_0$, and use two different loads $R_1$ and $R_2$ with
  

  $$I_1=\frac{V_0}{R_0+R_1},\;\;\;\;V_1=V_0\;\frac{R_1}{R_0+R_1},... ... I_2=\frac{V_0}{R_0+R_2},\;\;\;\;V_2=V_0\;\frac{R_2}{R_0+R_2}$$
  
  
  and
  
  $$\triangle I=I_1-I_2=V_0\;(\frac{1}{R_0+R_1}-\frac{1}{R_0+R_2}... ...angle V=V_1-V_2=V_0\;(\frac{R_1}{R_0+R_1}-\frac{R_2}{R_0+R_2})$$
  
  
  The output resistance can therefore be found to be $R_0$ as expected:
  
  $$-\frac{\triangle V}{\triangle I}=-\frac{V_1-V_2}{I_1-I_2}=R_0$$
  
  

  Consider two extreme cases for the two loads $R_1$ and $R_2$:

  • when $R_1=0$ (short circuit), then we get the short-circuit current:
    
    $$I_1=I_{sc}=V_0/R_0,\;\;\;\;\;\; V_1=0$$
    
    

  • when $R_2=\infty$ (open circuit), then we get the open-circuit voltage:
    
    $$V_2=V_{oc}=V_0,\;\;\;\;\;\;\;I_2=0$$
    
    

  and we have $\triangle V=V_1-V_2=-V_0$, $\triangle I=V_1-V_2=V_0/R_0$, and the output resistance is found to be:
  
  $$R_0=-\frac{\triangle V}{\triangle I}=\frac{V_{oc}}{I_{sc}}$$
  
  

  Example:

  

  • Thevenin's equivalent circuit:
    • Find open-circuit voltage $V_T=V_{oc}=V_{ab}=V_0-I_0R_1$
    • Find equivalent internal resistance $R_T=R_1$:
    • The Thevenin's equivalent circuit is shown in (b).
  • Norton's equivalent circuit:
    • Find short-circuit current. As there are two sources, superposition principle is used to get $I_N=I_{sc}=I_{ab}=V_0/R_1-I_0$
    • Find equivalent internal resistance $R_N=R_1$:
    • The Norton's equivalent circuit is shown in (c).
  Note that the resistor $R_2$ does not appear in either of the equivalent circuit. This is because $R_2$ is in series with an ideal current source which drives a constant current $I_0$ through the branch, independent of the resistance along the branch.

  Also note that the Thevenin's voltage source and the Norton's current source can be converted into each other:

  • Convert Thevenin's voltage source to Norton's current source:
    
    $$I_N=V_T/R_T=V_0/R_1-I_0$$
    
    

  • Convert Norton's current source to Thevenin's voltage source:
    
    $$V_T=I_N R_N=V_0-I_0R_1$$
    
    

• $\Delta$-Y Transformation

  

  The $\Delta$ configuration can be converted to $Y$ and vice versa. To relate $\Delta$ and $Y$, the resistances $R'_{ab}$ between terminals a and b of $Y$ should be equal to $R_{ab}$ of $\Delta$, and the same is true for the other two resistances, i.e.,
  

  $$\left\{ \begin{array}{rr} R'_{ab}=R_a+R_b=R_{ab}//(R_{ac}+R_... ... R'_{bc}=R_b+R_c=R_{bc}//(R_{ab}+R_{ac}) \end{array} \right.$$
  
  
  • Convert $\Delta$ to $Y$:

    Given $R_{ab}$, $R_{ac}$ and $R_{bc}$ of a $\Delta$, the three equations can be solved for $R_a$, $R_b$ and $R_c$ of the corresponding $Y$. For example, subtracting the 3rd equation from the sum of the first two, we get expression for $R_a$. The solutions are:
    

    $$\left\{ \begin{array}{rr} R_a=R_{ab}R_{ac}/(R_{ab}+R_{ac}+R_... ...c=R_{ac}R_{bc}/(R_{ab}+R_{ac}+R_{bc}) \\ \end{array} \right.$$
    
    

  • Convert $Y$ to $\Delta$:

    Reversely, given $R_a$, $R_b$ and $R_c$ of a $Y$, the same three equations can also be solved for $R_{ab}$, $R_{ac}$ and $R_{bc}$ of the corresponding $\Delta$ to get:
    

    $$\left\{ \begin{array}{rr} R_{ab}=R_a+R_b+R_aR_b/R_c \\ R_{... ...+R_aR_c/R_b \\ R_{bc}=R_b+R_c+R_bR_c/R_a \end{array} \right.$$
    
    
    The following circuit can be converted into either of the two low equivalent circuits.

  

  The $\Delta$ formed by $Z_1$, $Z_1$, and $Z_3$ can be converted to a Y, which can then be combined with $Z_4$ to get a Y (bottom left) with:
  

  $$\left\{ \begin{array}{l} Y_1=Z_1Z_3/(Z_1+Z_2+Z_3) \\ Y_2=Z... ...Z_3) \\ Y_3=Z_1Z_2/(Z_1+Z_2+Z_3)+Z_4 \\ \end{array} \right.$$
  
  
  Alternatively, the Y formed by $Z_1$, $Z_2$, and $Z_4$ can be converted to a $\Delta$, which can then be combined with $Z_3$ to get a $\Delta$ (bottom right) with:
  
  $$\left\{ \begin{array}{l} X_1=Z_1+Z_4+Z_1Z_4/Z_2 \\ X_2=Z_2... ..._3=(Z_1+Z_2+Z_1Z_2/Z_4) \vert\vert Z_3 \\ \end{array} \right.$$
  
  
  The resulting $\Delta$ and Y circuits are equivalent as it can be shown they can also be converted to each other with the same system variables.

  Example 0: (Homework)

  The conversion from $\Delta$ to $Y$ is more useful as $Y$ is easier to analyze than $\Delta$. For example, the circuit in (a) below can be converted to that in (b) to find all the currents in the circuit:

  

  Assume $V_0=225 V$, $R_0=1\Omega$, $R_1=40\Omega$, $R_2=36\Omega$, $R_3=50\Omega$, $R_4=55\Omega$, $R_5=10\Omega$, find the currents $I_0$, $I_1$, $I_2$, $I_3$, $I_4$, and $I_5$.

Example 1:

In the circuit below, $V_0=18V$, $R_1=R_2=3\Omega$, $R_3=6\Omega$, $R_4=1.5\Omega$. Find the value of current $I$ when $R_5$ is $1\Omega$, $2\Omega$, and $3\Omega$. Moreover, find the value for $R_5$ for the desired current $I=0.5 A$.

Method 1, $\Delta-Y$ conversion

Find $I$ when $R_5=2\Omega$. First convert the $\Delta$ composed of $R_1$, $R_2$ and $R_5$ into a $Y$ composed of $R_a$, $R_b$ and $R_c$:

$$R_c=\frac{R_1R_2}{R_1+R_2+R_5}=\frac{9}{8},\;\;\;\; R_a=\frac{R_1R_5}{R_1+R_2+R_5}=\frac{3}{4}=R_b$$

Find overall resistance:

$$R_{total}=R_c+(R_a+R_3) \vert\vert (R_b+R_4)=\frac{45}{16}$$

Find overall current:

$$I_0=\frac{V_0}{R_{total}}=\frac{18}{45/16}=\frac{32}{5}$$

Find currents through $R_3$ and $R_4$ (current divider):

$$I_a=I_0\;\frac{R_b+R_4}{(R_a+R_3)+(R_b+R_4)}=\frac{8}{5}$$

$$I_b=I_0\;\frac{R_a+R_3}{(R_a+R_3)+(R_b+R_4)}=\frac{24}{5}$$

Find voltage at points $a$ and $b$ (assuming negative end of voltage source is ground):

$$V_a=I_a \times R_3=\frac{8 }{5} \times 6=\frac{48}{5},\;\;\;\;\; V_b=I_b \times R_4=\frac{24}{5} \times 1.5=\frac{36}{5}$$

Find current $I$ through $R_5=2\Omega$:

$$I=\frac{V_a-V_b}{R_5}=\frac{12}{5}\times \frac{1}{2}=1.2$$

The same steps can be repeated for $R_5=1\Omega$ and $R_5=3\Omega$. But it is hard to find a value of $R_5$ given the require current $I=0.5$.

Method 2, Thevenin's theorem

Solve the problem using Thevenin's theorem by the following steps:

• remove the branch in question from the circuit and treat the rest as a one-port network.
• simplify the one-port network by Thevenin's theorem, find the open circuit voltage $V_T$ and the equivalent internal resistance $R_T$.
• put the branch in equation back as the load of the Thevenin equivalent network and find the current/voltage.

Here, we remove $R_5$ as the load of a network composed of all other resistors $R_1$, $R_2$, $R_3$, $R_4$ and the voltage source $V_0=18V$, then apply Thevenin's theorem to find the open-circuit voltage between the two terminals a and b:

$$V_T=V_{oc}=V_0\frac{R_3}{R_1+R_3}-V_0\frac{R_4}{R_2+R_4} =18 (\frac{6}{9}-\frac{1.5}{4.5})=6V$$

and the internal resistance between a and b (with voltage source $V_0$ short circuit):

$$R_T=R_1//R_3+R_2//R_4=\frac{3\times 6}{3+6}+\frac{3\times 1.5}{3+1.5}=3\Omega$$

Now find current $I$ for different $R_5$

• $R_5=1$, $I=6/(3+1)=1.5$
• $R_5=2$, $I=6/(3+2)=1.2$
• $R_5=3$, $I=6/(3+3)=1.0$

and when $I=0.5$, $R_5=V_T/I-R_T=6/0.5-3=9$

Example 2: The circuit below, often used in some control system, is composed of two voltages, two potentiometers, and a load resistor. Assume $V_1=72V$, $V_2=80V$, $R_1=1.5\Omega$, $R_2=3\Omega$, $R_3=1.5\Omega$, and $R_4=2.5\Omega$. Find the current $I_L$ through the load resistor $R_L=1.5\Omega$.

Method 1, Superposition theorem

Find $I'_L$ caused by voltage $V_1=72$, and then $I''_L$ caused by voltage $V_2=80$, then get $I_L=I'_L+I''_L$.

• Short circuit $V_2$. Assume $I'_L=8$, so that currents through $R_3$ and $R_4$ are, respectively, $I_3=5$ and $I_4=3$ (current divider), and $V_b=7.5$.
• $V_a=R_L I'_L+V_b=1.5\times 8+7.5=19.5$, current through $R_2$ is $I_2=V_a/R_2=19.5/3=6.5$, current through $R_1$ is $I_1=I_2+I'_L=6.5+8=14.5$.
• $V'_1=R_1\times I_1+V_a=1.5\times 14.5+19.5=41.25$. But $V_1=72$, we get scaling factor $V_1/V'_1=72/41.25=96/55$, and $I'_L=96\times 8/55$.

• Short circuit $V_1$. Assume $I''_L=3$, so that currents through $R_1$ and $R_2$ are, respectively, $I_1=2$ and $I_2=1$ (current divider), and $V_a=3$.
• $V_b=R_L I''_L+V_a=1.5\times 3+3=7.5$, current through $R_3$ is $I_3=V_b/R_3=7.5/1.5=5$, current through $R_4$ is $I_4=I_3+I''_L=5+3=8$.
• $V'_2=R_4\times I_4+V_b=2.5\times 8+7.5=27.5$. But $V_2=80$, we get scaling factor $V_2/V'_2=80/27.5=32/11$, and $I''_L=3\times 32/11=96/11$.

• Finally, we get the load current:
  
  $$I_L=I'_L-I''_L=\frac{96\times 8}{55}-\frac{96}{11}=\frac{288}{55}$$
  
  

Method 2, Thevenin's theorem

Remove $R$, find open-circuit voltage $V_{ab}=V_a-V_b$ and equivalent resistance $R$, then find $I_L=V_{ab}/(R+R_L)$.

• $V_a=V_1\times R_2/(R_1+R_2)=72\times 3/(1.5+3)=48$

  $V_b=V_2\times R_3/(R_3+R_4)=80\times 1.5/(1.5+2.5)=30$

  $V_T=V_{ab}=V_a-V_b=48-30=18$

• $R_T=R_1 \vert\vert R_2 + R_3 \vert\vert R_4=3\vert\vert 1.5+2.5\vert\vert 1.5=7.75/4$
• 
  
  $$I_L=\frac{V_{ab}}{R+R_L}=\frac{18}{1.5+7.75/4}=\frac{288}{55}$$