Circuit Theorems

• Superposition Theorem:

  A generic system with input (stimulus, cause) $x$ and output (response, effect) $y$ can be described mathematically as a function $y=F(x)$. The function is linear if it satisfies the following:

  • Homogeneity

    $$F(ax)=aF(x)$$ (19)

  • Superposition

    $$F(x_1+x_2)=F(x_1)+F(x_2)$$ (20)

  Combining these two aspects, we have

  $$F(ax_1+bx_2)=aF(x_1)+bF(x_2)$$ (21)

  The interpretation is that when the response is directly proportional to the cause in a system $F$, we can consider several causes (e.g., $x_1$ and $x_2$) individually and then combine the individual responses ($F(x_1)$ and $F(x_2)$). An electrical system of linear components (with linear voltage-current relation) is a linear system (if only voltage and current are of interest). When the circuit is linear, the following applies:

  Superposition Principle:

  When there exist multiple energy sources in the circuit, any voltage and current 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:

  • A voltage source is turned off if treated as short-circuit so that the voltage across it is guaranteed to be zero.
  • A current source is turned off if treated as open-circuit so that the current through it is guaranteed to be zero.

  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:

    $$I_{12}=\frac{aV_1+bV_2}{R}=a\frac{V_1}{R}+b\frac{V_2}{R}=aI_1+bI_2$$ (22)

    where $I_1=V_1/R$ ($V_2$ short-circuit) and $I_2=V_2/R$ ($V_1$ short-circuit).

  • Superposition of current sources:

    $$V=R(aI_1+bI_2)=aI_1R+bI_2R=aV_1+bV_2$$ (23)

    where $V_1=I_1R$ ($I_2$ open circuit) and $V_2=I_2R$ ($I_1$ open circuit).

  However, note that superposition principle does not apply to any variable nonlinearly related to the energy sources, such as power:

  $$P_{12}=\frac{(V_1+V_2)^2}{R}\ne\frac{V_1^2}{R}+\frac{V_2^2}{R}=P_1+P_2, \;\;\;\;\;\;\; P_{12}=R(I_1+I_2)^2 \ne RI_1^2+RI_2^2=P_1+P_2$$ (24)

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

  

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

  $$I'_1=\frac{32}{2+8 \vert\vert 4}=\frac{48}{7},\;\;\;I'_2=-I'_1\frac{8}{8+4}=-\frac{32}{7}, \;\;\;I'_3=I'_1\frac{4}{8+4}=\frac{16}{7}$$ (25)

  Second, we turn the voltage source of 32V off and get

  $$I''_2=\frac{20}{4+8 \vert\vert 2}=\frac{25}{7},\;\;\;I''_1=-I''_2\frac{8}{2+8}=-\frac{20}{7}, \;\;\;I''_3=I''_2\frac{2}{8+2}=\frac{5}{7}$$ (26)

  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:

  $$I_1=I'_1+I''_1=\frac{48}{7}-\frac{20}{7}=4,\;\;\;\; I_2=I''_2+I'_2=\frac{25}{7}-\frac{32}{7}=-1,\;\;\;\; I_3=I'_3+I''_3=\frac{16}{7}+\frac{5}{7}=3$$ (27)

  Example 2: Find the voltage across the parallel combination of two branches $R_1=3\,\Omega$ in series with $V_1=12\,V$ and $R_2=6\,\Omega$ in series with $V_2=6\,$.

  

  This problem can be solved by different methods:

  • The total corrent is

    $$I=\frac{V_1-V_2}{R_1+R_2}=\frac{12-6}{3+6}=2/3$$ (28)

    The voltages accross $R_1$ and $R_2$ are

    $$V_{R_1}=R_1I=2\,V,\;\;\;\;\;\;V_{R_2}=R_2I=4\,V$$ (29)

    The total voltage is $V=V_1-V_{R_1}=V_2+V_{R_2}=10\,V$.

  • Convert the two branches of non-ideal voltage sources into two non-ideal current sources:

    $$I_1=V_1/R_1=12/3=4,\;\;\;\;\;\;\;\;I_2=V_2/R_2=6/6=1$$ (30)

    The total current is $I=I_1+I_2=5\,A$, and the total resistance is $R=R_1 R_2/(R_1+R_2)=18/9=2\,\Omega$. This current source can be converted into a voltage source with $V=10\,V$ and $R=2\,\Omega$. In either case, we get the voltage $5\times 2=10\,V$.

  • First find voltages due to $V_1=12$ and $V_2$ alone:

    $$V'=\frac{R_2}{R_1+R_2}V_1=\frac{6}{3+6}\,12=8\,V,\;\;\;\;\; V''=\frac{R_1}{R_1+R_2}V_2=\frac{3}{3+6}\,6=2\,V$$ (31)

    Then based on the superposition theorem, we get $V=V'+V''=10\,V$.

  Thevenin's theorem and Norton's theorem

  In principle, all currents and voltages of an arbitrary network of linear components and voltage/current sources can be found by either the loop current method or the node voltage method, as we have seen previously.

  However, if only the current $I$ and voltage $V$ associated with one particular component such as a resistor $R$ are of interest, it is unnecessary to find voltages and currents elsewhere in the circuit. Instead, we can “pull” the component out and treat it as the load $R_L$ of the rest of the circuit, which can be modeled as either a Thevenin voltage source (Leon Charles Thevenin), a non-ideal voltage source $(V_T, R_T)$, or a Norton current source (Edward Lawry Norton) a non-ideal current source $(I_N, R_N)$. We can then find the $V$ and $I$ associated with the resistor $R$.

  

• Thevenin's Theorem:

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

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

  Proof:

  

  • Assume the voltage and current associated with the load $R_L$ at the output port of the network are $V$ and $I$, respectively.

  • Replace the load $R_L$ in circuit (a) by an ideal current source $I=V/R_L$ while keeping voltage $V$ the same (b), all voltages and currents in the network to be modeled are not affected.

  • Find voltage $V$ in terms of the internal energy sources inside the network and the external current source by superposition principle:
    • Turn the external current source off (open-circuit), the terminal voltage is the open-circuit voltage $V'=V_{oc}$ due only to the energy sources internal to the network (c).
    • Turn all internal sources off (short-circuit for voltage sources, open-circuit for current sources), the terminal voltage is $V''=-R_0I$, where $R_0$ is the equivalent resistance of the network with all energy sources off, due only to the external current source (d).
    • By the superposition principle, we get the load voltage $V=V'+V''=V_{oc}-R_0I$.
    • The voltage $V$ and current $I$ in the circuit in (e) are also related by the same equation $V=V_T-R_TI$, i.e., this circuit is equivalent to the original circuit in (a), and can therefore be used as a model of the circuit.

  We see that as far as the port voltage $V$ and current $I$ associated with the load are concerned, the one-port network is equivalent to an ideal voltage source $V_T=V_{oc}$, 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 (can be modeled by) an ideal current source $I_N$ in parallel with a resistor $R_N$, where

  • $I_N=I_{sc}$ is the short-circuit current of the network
  • $R_N=R_T$ is the same resistance as in Thevenin's theorem.

  Proof: The proof of this theorem is in parallel with the proof of Thevenin's theorem.

  

  • Assume the voltage and current associated with the load $R_L$ at the output port of the network are $V$ and $I$, respectively.

  • Replace the load $R_L$ in circuit (a) by an ideal voltage source $V=IR_L$ while keeping current $I$ the same (b), all voltages and currents in the network to be modeled are not affected.

  • Find current $I$ in terms of the internal energy sources inside the network and the external voltage source by superposition principle:
    • Turn the external voltage source off (short-circuit), the terminal current is the short-circuit current $I'=I_{sc}$ due only to the energy sources internal to the network (c).
    • Turn all internal sources off (short-circuit for voltage sources, open-circuit for current sources), the terminal current is $I''=-V/R_0$, where $R_0$ is the equivalent resistance of the network with all energy sources off, due only to the external voltage source (d).
    • By the superposition principle, we get the load current $I=I'+I''=I_{sc}-V/R_0$.
    • The current $I$ and voltage $V$ in the circuit in (e) are also related by the same equation $I=I_N-V/R_N$, i.e., this circuit is equivalent to the original circuit in (a), and can therefore be used as a model of the circuit.

  We see that as far as the port voltage $V$ and current $I$ associated with the load are concerned, the one-port network is equivalent to an ideal current source $I_N=I_{sc}$, 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.

  Moreover, we note that Thevenin's theorem and Norton's theorem are equivalent, as one can always be converted into the other. The internal resistances in both theorems are the same $R_T=R_N$, and the voltage source $V_T=V_{oc}$ in series with $R_T$ in Thevenin's theorem can be converted to a current source $V_{oc}/R_T=V_T/R_T=I_{sc}=I_N$ in parallel with $R_N=R_T$ in Norton's theorem. In either case, we can find the internal resistance by

  $$R_T=R_N=\frac{V_{oc}}{I_{sc}}$$ (32)

• Millman's Theorem:

  The voltage across a set of $N$ parallel branches of voltage source $V_i$ in series with $R_i$ ( $i=1,\cdots,n$) is

  $$V=\frac{\sum_{i=1}^n \frac{V_i}{R_i}}{\sum_{i=1}^n\frac{1}{R_i}}$$ (33)

  Proof: (Homework)

  

  This theorem can be augmented to include branches containing $R_j$ but no voltage source ($V_j=0$), and branches containing a current source $I_j$. If a resistance $R_j$ is in series with $I_j$, it can be neglected (as it does not affect the current). If a resistance $R_j$ is in parallel with $I_j$, they can be converted into a voltage source $V_j=I_jR_j$ in series with $R_j$.

• $\Delta$-Y Transformation

  

  The $\Delta$ configuration can be converted to $Y$ and vice versa. We relate the $\Delta$ and $Y$ by realizing that the resistance $R'_{ab}$ between terminals a and b of $Y$ should be equal to that between the same two terminals of $\Delta$:

  $$\left\{ \begin{array}{rr} R'_{ab}=R_a+R_b=R_{ab}\vert\vert(R_{ac}... ...R_{bc}) \\ R'_{bc}=R_b+R_c=R_{bc}\vert\vert(R_{ab}+R_{ac}) \end{array} \right.$$ (34)

  • Convert $\Delta$ to $Y$:

    Given $R_{ab}$, $R_{ac}$ and $R_{bc}$ of a $\Delta$, the three equations above 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:

    $$\left\{ \begin{array}{rr} R_a=R_{ab}R_{ac}/(R_{ab}+R_{ac}+R_{bc})... ...ac}+R_{bc}) \\ R_c=R_{ac}R_{bc}/(R_{ab}+R_{ac}+R_{bc}) \\ \end{array} \right.$$ (35)

  • Convert $Y$ to $\Delta$:

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

    $$\left\{ \begin{array}{rr} R_{ab}=R_a+R_b+R_aR_b/R_c \\ R_{ac}=R_a+R_c+R_aR_c/R_b \\ R_{bc}=R_b+R_c+R_bR_c/R_a \end{array} \right.$$ (36)

  The top circuit (a bridged T-network) in the following figure can be converted into either of the two equivalent circuits below.

  

  • 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):

    $$\left\{ \begin{array}{l} Y_1=Z_1Z_3/(Z_1+Z_2+Z_3) \\ Y_2=Z_2Z_3/(Z_1+Z_2+Z_3) \\ Y_3=Z_1Z_2/(Z_1+Z_2+Z_3)+Z_4 \\ \end{array} \right.$$ (37)

  • 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):

    $$\left\{ \begin{array}{l} X_1=Z_1+Z_4+Z_1Z_4/Z_2 \\ X_2=Z_2+Z_4+Z_2Z_4/Z_1 \\ X_3=(Z_1+Z_2+Z_1Z_2/Z_4) \vert\vert Z_3 \\ \end{array} \right.$$ (38)

  The resulting $\Delta$ and Y circuits are equivalent as it can be shown that they can also be converted to each other with the same system variables.

Example 1: Model the circuit in part (a) by Thevenin's theorem (b) and Norton's theorem (c).

Find equivalent internal resistance when both energy sources are turned off: $R_T=R_N=R_1$.

• Thevenin's equivalent circuit (b):

  Find open-circuit voltage: $V_T=V_{oc}=V_{ab}=V_0+I_0R_1$

• Norton's equivalent circuit (c):

  Find short-circuit current (superposition): $I_N=I_{sc}=I_{ab}=V_0/R_1+I_0$

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

$$I_N=V_T/R_T=V_0/R_1+I_0,\;\;\;\;\;\;\;\;\; V_T=I_N R_N=V_0+I_0R_1$$ (39)

Also note that the resistors $R_0$ and $R_2$ do not appear in either of the two equivalent circuits, because a voltage source provides a constant voltage $V_0$ independent of any resistance in parallel, and a current source drives a constant current $I_0$, independent of any resistance in series.

Example 2: Find voltage $V$ across and current $I$ through $R_2$.

Any one of the following methods can be used to solve the circuit:

• Loop-current method:

  Apply KVL around the outer loop with loop current $I$ to get

  $$I R_2+(I_0+I) R_1-V_0=0$$ (40)

  Solving this we get

  $$I=\frac{V_0-I_0 R_1}{R_1+R_2}$$ (41)

• Node-voltage method:

  Assume the currents $I_1$ (left branch), $I_0$ and $I_2=I$ (right branch) all leave the top node, where the voltage is $V$ (with respect to the bottom treated as ground). By KCL, we have

  $$I_0+I_1+I_2=I_0+\frac{V-V_0}{R_1}+\frac{V}{R_2}=0,\;\;\;\; \;\;\;\; I_0R_1R_2+R_2(V-V_0)+R_1V=0$$ (42)

  Solving for $V$, we get:

  $$V=\frac{R_2V_0-I_0R_1R_2}{R_1+R_2} =V_0\frac{R_2}{R_1+R_2}-I_0\frac{R_1R_2}{R_1+R_2}$$ (43)

  and

  $$I=I_2=\frac{V}{R_2}=\frac{V_0}{R_1+R_2}-I_0\frac{R_1}{R_1+R_2}$$ (44)

• Superposition principle:

  

  • Find $V'$ and $I'$ with the current source off (open-circuit with zero current):

    $$I'=\frac{V_0}{R_1+R_2},\;\;\;\;V'=I'R_2=V_0\frac{R_2}{R_1+R_2}$$ (45)

  • Find $V''$ and $I''$ with the voltage source off (short-circuit with zero voltage):

    $$I''=-I_0\frac{R_1}{R_1+R_2},\;\;\;\;V''=I''R_2=-I_0\frac{R_1R_2}{R_1+R_2}$$ (46)

    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 partial results we get

    $$I=I'+I''=\frac{V_0}{R_1+R_2}-I_0\;\frac{R_1}{R_1+R_2}$$ (47)

    $$V=V'+V''=V_0\;\frac{R_2}{R_1+R_2}-I_0\;\frac{R_1R_2}{R_1+R_2}$$ (48)

• Energy source conversion:
  • Convert $V_0$ and $R_1$ in the left branch as a non-ideal voltage source into a current source $I_1=V_0/R_1$ (upward);
  • Combine the two parallel current sources in parallel with two resistors $R_1$ and $R_2$;
  • find $I$ (current devider):

    $$I=\left(\frac{V_0}{R_1}-I_0\right)\frac{R_1}{R_1+R_2} =\frac{V_0}{R_1+R_2}-I_0\frac{R_1}{R_1+R_2}$$ (49)

  • Find $V$:

    $$V=R_2I=V_0\frac{R_2}{R_1+R_2}-I_0\frac{R_1R_2}{R_1+R_2}$$ (50)

• Thevenin's and Nordon's methods:
  • Thevenin's model: $R_T=R_1,\;\;V_T=V_0-I_0R_1$,

    $$I=\frac{V_T}{R_T+R_2}=\frac{V_0-I_0R_1}{R_1+R_2} =\frac{V_0}{R_1+R_2}-I_0\frac{R_1}{R_1+R_2}$$ (51)

  • Norton's model: $R_N=R_1,\;\;I_N=V_0/R_1-I_0$.

    $$I=\left(\frac{V_T}{R_1}-I_0\right)\frac{R_N}{R_N+R_2} =\frac{V_0}{R_1+R_2}-I_0\frac{R_1}{R_1+R_2}$$ (52)

  Then we get the voltage:

  $$V=R_2I=V_0\frac{R_2}{R_1+R_2}-I_0\frac{R_1R_2}{R_1+R_2}$$ (53)

Example 3: 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 $R_5=2\Omega$. Then re-solve the problem when $R_5$ takes different values of $1\Omega$ and $3\Omega$. Moreover, find the value of $R_5$ so that $I=0.5 A$ as desired.

• Loop-current (KVL) method:

  Consider three independent loops with clockwise loop currents:

  • $I_0$ in the left loop ($V_0$, $R_2$, $R_4$)
  • $I_1$ in the top loop ($R_2$,$R_1$, $R_5$)
  • $I_2$ in the bottom loop ($R_3$, $R_4$, $R_5$)
  Applying KVL to the three loops we get three simultaneous equations:

  $$\left\{\begin{array}{l} R_2(I_0-I_1)+R_4(I_0-I_2)=3(I_0-I_1)+1.5(... ...I_2+R_4(I_2-I_0)+R_5(I_2-I_1)=6I_2+1.5(I_2-I_0)+2(I_2-I_1) =0\end{array}\right.$$ (54)

  Rearranging and solving the equations we get

  $$\left\{\begin{array}{l} 3I_0-2I_1-I_2=12\\ -3I_0+8I_1-2I_2=0\\ 3I... ...tarrow \left\{\begin{array}{l} I_0=32/5\\ I_1=14/5\\ I_2=8/5 \end{array}\right.$$ (55)

  The node voltages can be found to be

  $$V_a=R_3I_2=6\times \frac{8}{5}=\frac{48}{5},\;\;\;\;\;\; V_b=R_4(... ...2)=1.5\times \frac{24}{5}=\frac{36}{5},\;\;\;\;\;\; V_{ab}=V_a-V_b=\frac{12}{5}$$ (56)

  The current through $R_5$ is

  $$I_1-I_2=\frac{14}{5}-\frac{8}{5}=\frac{6}{5}=1.2\;A, \;\;\;\;\;\;\; \;\;\;\;\;\; I=\frac{V_a-V_b}{R_5}=\frac{12/5\;V}{2\;\Omega}=1.2\;A$$ (57)

  We have to repeat the above for other values of $R_5$ to find the current through it.

• Node-Voltage (KCL) method:

  Assume the bottom node is ground with 0 volt, then the voltage at the top node is known to be $V_0=18\,V$. Now we apply KCL to nodes a and b to get two equations:

  $$\frac{V_a-V_0}{R_1}+\frac{V_a-V_b}{R_5}+\frac{V_a}{R_3} =\frac{V_a-18}{3}+\frac{V_a-V_b}{2}+\frac{V_a}{6}=0$$ (58)

  $$\frac{V_b-V_0}{R_2}+\frac{V_b-V_a}{R_5}+\frac{V_b}{R_4} =\frac{V_b-18}{3}+\frac{V_b-V_a}{2}+\frac{V_b}{1.5}=0$$ (59)

  $$\left\{ \begin{array}{ll}-V_a+3V_b=12\\ 2V_a-V_b=12\end{array}\ri... ...rrow\;\;\;\;\;\; \left\{ \begin{array}{ll}V_a=48/5\\ V_b=36/5\end{array}\right.$$ (60)

  The current through $R_5$ is

  $$I=\frac{V_a-V_b}{R_5}=\frac{(48/5-36/5)\,V}{2\,\Omega} =\frac{12/5\;V}{2\;\Omega}=\frac{6}{5}=1.2\;A$$ (61)

  We have to repeat the above for other values of $R_5$ to find the current through it.

  To find $R_5$ for the current through it to be $I=0.5$, we replace the current $I=(V_a-V_b)/2$ through $R_5$ in the two equations above by the desired $I=0.5$:

  $$\frac{V_a-18}{3}+0.5+\frac{V_a}{6}=0,\;\;\;\;\;\; \frac{V_b-18}{3}-0.5+\frac{V_b}{1.5}=0$$ (62)

  Solving each of the two equations we get $V_a=11\,V$, $V_b=6.5\,V$, and

  $$R_5=\frac{V_a-V_b}{I}=\frac{11-6.5}{0.5}=9\;\Omega$$ (63)

• $\Delta-Y$ conversion

  

  To find $I$ through $R_5=2\Omega$, we first convert the $\Delta$ composed of $R_1$, $R_2$ and $R_5$ into a $Y$ composed of $R_a$, $R_b$ and $R_c$:

  $$R_c=\frac{R_1R_2}{R_1+R_2+R_5}=\frac{9}{8},\;\;\;\; R_a=\frac{R_1R_5}{R_1+R_2+R_5}=\frac{3}{4},\;\;\;\; R_b=\frac{R_2R_5}{R_1+R_2+R_5}=\frac{3}{4}$$ (64)

  Find overall load resistance of $V_0=18\,V$:

  $$R_0=R_c+(R_a+R_3) \;\vert\vert\; (R_b+R_4)=\frac{45}{16}$$ (65)

  Find overall current:

  $$I_0=\frac{V_0}{R_0}=\frac{18}{45/16}=\frac{32}{5}$$ (66)

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

  $$I_a=I_0\;\frac{R_b+R_4}{(R_a+R_3)+(R_b+R_4)}=\frac{8}{5}$$ (67)

  $$I_b=I_0\;\frac{R_a+R_3}{(R_a+R_3)+(R_b+R_4)}=\frac{24}{5}$$ (68)

  Find voltage at points $a$ and $b$ (assuming the bottom node is ground):

  $$V_a=I_a \times R_3=\frac{8 }{5} \times 6=\frac{48}{5},\;\;\;\;\; V_b=I_b \times R_4=\frac{24}{5} \times 1.5=\frac{36}{5}$$ (69)

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

  $$I=\frac{V_a-V_b}{R_5}=\frac{12}{5}\times \frac{1}{2}=1.2A$$ (70)

  For any of the methods above, the problem needs to be resolved for $R_5=1\Omega$ and $R_5=3\Omega$, and it is hard to find a value of $R_5$ given the require current $I=0.5$.

  

• 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.
  • Model the one-port network by Thevenin's theorem, as an open circuit voltage $V_T$ in series with an internal resistance $R_T$.
  • Put the branch in equation back as the load of the Thevenin equivalent network and find the desired current and voltage.

  Specifically, 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:

  $$V_T=V_{oc}=V_0\frac{R_3}{R_1+R_3}-V_0\frac{R_4}{R_2+R_4} =18 \left(\frac{6}{9}-\frac{1.5}{4.5}\right)=6V$$ (71)

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

  $$R_T=R_1\,\vert\vert\,R_3+R_2\,\vert\vert\,R_4 =\frac{3\times 6}{3+6}+\frac{3\times 1.5}{3+1.5}=3\Omega$$ (72)

  Now the currents through $R_5$ of different values can be found by

  $$I=\frac{V_T}{R_T+R_5}$$ (73)

  • $R_5=1$, $I=6/(3+1)=1.5A$
  • $R_5=2$, $I=6/(3+2)=1.2A$
  • $R_5=3$, $I=6/(3+3)=1.0A$
  and when $I=0.5$, $R_5=V_T/I-R_T=6/0.5-3=9$

Example 4: The circuit below, often used in some control system, is composed of two voltages, two potentiometers, and a load resistor. Assume:

$$V_1=72V,\;\;\;\;V_2=80V,\;\;\;\;R_1=R_3=1.5\Omega,\;\;\;\;R_2=3\Omega, \;\;\;\;R_4=2.5\Omega$$ (74)

Find the current $I_L$ through the load resistor $R_L=1.5\Omega$.

We denote the current through $R_i$ by $I_i$, and the voltage at the left and right ends of $R_L$ by $V_a$ and $V_b$, respectively, with respect to the bottom wire treated as the ground.

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 $R_3=1.5$ and $R_4=2.5$ are $I_3=5$ and $I_4=3$, respectively (current divider), and $V_b=I_3R_3=I_4R_4=5\times 1.5=3\times 2.5=7.5$.
• $V_a=R_L I'_L+V_b=1.5\times 8+7.5=19.5$, current through $R_2=3$ 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 $I'_L=8 \times 96/55$.

• Short-circuit $V_1$. Assume $I''_L=3$, so that currents through $R_1=1.5$ and $R_2=3$ are, respectively, $I_1=2$ and $I_2=1$ (current divider), and $V_a=R_1I_1=R_2I_2=1.5\times 2+3\times 1=3$.
• $V_b=R_L I''_L+V_a=1.5\times 3+3=7.5$, current through $R_3=1.5$ 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:

  $$I_L=I'_L-I''_L=\frac{96\times 8}{55}-\frac{96}{11}=\frac{288}{55}$$ (75)

Thevenin's theorem

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

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

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

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

• $R_T=R_1 \vert\vert R_2 + R_3 \vert\vert R_4=3\vert\vert 1.5+2.5\vert\vert 1.5=7.75/4$

• $$I_L=\frac{V_T}{R_T+R_L}=\frac{18}{7.75/4+1.5}=\frac{288}{55}$$ (76)

Example 5: In the circuit below, $R_1=6\Omega$, $R_2=2\Omega$, $R_3=3\Omega$, $R_4=5\Omega$, $R_5=2\Omega$, $V=5V$. Given the current (downward) through $R$ is $I=0.5 A$, find the resistance of $R$.

Node voltage method:

Let the voltages at the middle node be $V_a$ and at the top-right node be $V_b$. Then based on node-voltage method, we get

$$\frac{V_a-5}{2}+\frac{V_a}{2}+\frac{V_a-V_b}{3}=0,\;\;\;\;\; \frac{V_b-V_a}{3}+\frac{V_b-5}{6}+0.5=0$$ (77)

Solving these equations we get $V_a=2.45$, $V_b=2.3V$, and $R=V_b/I=2.3/0.5=4.6\Omega$.

Thevenin's theorem: (Homework)

Solution

Example 6: (Homework)

In the circuit below, $R_1=6\Omega$, $R_2=R_3=15\Omega$, $R_4=10\Omega$, $V_1=6V$, $V_2=9V$, $I_1=2A$ and $I_2=1A$. Find the Thevenin model of this circuit in terms of $V_T$ and $R_T$.

Solution