Solving Circuits with Kirchoff Laws

Example 1: Find the three unknown currents ( $I_1,I_2,I_3$) and three unknown voltages ( $V_{ab}, V_{bd}, V_{cb}$) in the circuit below:

Note: The direction of a current and the polarity of a voltage can be assumed arbitrarily. To determine the actual direction and polarity, the sign of the values also should be considered. For example, a current labeled in left-to-right direction with a negative value is actually flowing right-to-left.

All voltages and currents in the circuit can be found by either of the following two methods, based on KVL or KCL respectively.

• The loop-current method (mesh current analysis) based on KVL:
  1. For each of the $l$ independent loops in the circuit, define a loop current around the loop in clockwise (or counter clockwise) direction. These $l$ loop currents are the unknown variables to be obtained.
  2. Apply KVL around each of the loops in the same clockwise direction to obtain $l$ equations. While calculating the voltage drop across each resistor shared by two loops, both loop currents (in opposite positions) should be considered.
  3. Solve the equation system with $l$ equations for the $l$ unknown loop currents.

  

  Find currents $I_1$ from a to b, $I_2$ from c to b, and $I_3$ from b to d.

  • Assume two loop currents $I_a$ and $I_b$ around loops abda and bcdb and apply the KVL to them:
    

    $$\sum V_{abda} = -32+2I_a+8(I_a-I_b)=0$$

    $$\sum V_{bcdb} = 8(I_b-I_a)+4I_b+20=0$$ (1)

    
    We rewrite these as:

    $$\left\{ \begin{array}{rrrr} 10I_a&-8I_b&=&32 \\ -8I_a &+12I_b&=&-... ...ghtarrow \;\;\;\;\; \left\{ \begin{array}{l} I_a=4 \\ I_b=1 \end{array} \right.$$ (2)

    and then get $I_1=I_a=4$, $I_2=-I_b=-1$, and $I_3=I_a-I_b=3$. Having found $I_a$ and $I_b$, we can easily find all voltages in the circuit: $V_a=32V$, $V_b=32-2\times 4=24V$, $V_c=20V$, $V_{ab}=V_a-V_b=8\,V$, $V_{bd}=V_b-V_d=V_b=24\,V$, and $V_{cb}=V_c-V_b=-4$.

  • We could also apply KVL around the third loop of abcda with a loop current $I_c$ to get three equations:
    

    $$\sum V_{abda} = -32+2(I_a+I_c)+8(I_a-I_b)=0$$

    $$\sum V_{bcdb} = 8(I_b-I_a)+4(I_b+I_c)+20=0$$

    $$\sum V_{abcda} = -32+2(I_a+I_c)+4(I_b+I_c)+20=0$$ (3)

    
    However, it is clear that the third equation is the sum of the first two equations, i.e., it is not independent.

  • Alternatively, consider the two loop currents $I_a$ and $I_c$ around loops abda and bcdb:
    

    $$\sum V_{abda} = -32+2(I_a+I_c)+8I_a=0$$

    $$\sum V_{abcda} = -32+2(I_a+I_c)+4I_c+20=0$$ (4)

    
    i.e.,

    $$\left\{ \begin{array}{rrrr} 10I_a&+2I_c&=&32 \\ 2I_a &+6I_c&=&12 ... ...ghtarrow \;\;\;\;\; \left\{ \begin{array}{l} I_a=3 \\ I_c=1 \end{array} \right.$$ (5)

    and we get $I_1=I_a+I_c=4$ and $I_2=-I_c=-1$, same as the previous results.

• The node-voltage method (nodal voltage analysis) based on KCL:
  1. Assume there are $n$ nodes in the circuit. Select one of them as the ground, the reference point for all voltages of the circuit. The node voltage at each of the remaining $n-1$ nodes is an unknown to be obtained.
  2. Express each current into a node in terms of the two associated node voltages.
  3. Apply KCL to each of the $n-1$ nodes to set the sum of all currents into the node to zero, and get $n-1$ equations.
  4. Solve the equation system with $n-1$ equations for the $n-1$ unknown node voltages.

  

  In the same circuit considered previously, there are only 2 nodes $b$ and $d$ ($a$ and $c$ are not nodes). We assume node $d$ is the ground, and consider just voltage $V_b=V_{bd}$ at node $v$ as the only unknown in the problem. Apply KCL to node $b$, we have

  $$\sum_i I_i=0=I_1+I_2+I_3$$ (6)

  where each current is expressed as the voltage drop between the two ends of a resistor in the branch divided by the resistance of the resistor (Ohm's law):

  $$I_1=\frac{V_a-V_b}{R_{ab}}=\frac{32-V_b}{2},\;\;\;\; I_2=\frac{V_... ...b}{R_{cb}}=\frac{20-V_b}{4},\;\;\;\; I_3=\frac{V_d-V_b}{R_{bd}}=\frac{0-V_b}{8}$$ (7)

  Substituting $I_1$, $I_2$, and $I_3$ into the equation, we get

  $$I_1+I_2+I_3=\frac{32-V_b}{2}+\frac{20-V_b}{4}+\frac{0-V_b}{8}=0$$ (8)

  Solving this we get $V_b=24$, and all other currents and voltages can be found subsequently: $I_1=(V_a-V_b)/2=(32-24)/2=4A$, $I_2=(V_c-V_b)/4=(20-24)/4=-1A$, $I_3=I_1+I_2=3A$.

  We could also apply KCL to node d, but the resulting equation is exactly the same as $\sum_i I_i=0=-I_1-I_2-I_3$ simply because this node d is not independent.

  As special case of the node-voltage method with only two nodes, we have the following theorem:

• Millman's theorem

  If there are multiple parallel branches between two nodes $a$ and $b$, such as the circuit below (left), then the voltage $V$ at node $a$ can be found as shown below if the other node $b$ is treated as the reference point.

  Assume there are three types of branches:

  • voltage branches with $V_i$ sources in series with $R_i$. The polarity of each $V_i$ is + on the node a side.
  • current branches with $I_j$ (independent of resistors in series). The direction of each $I_j$ is toward node a.
  • resistor branches with $R_k$.

  Applying KCL to node $a$, we have:

  $$\sum_i \frac{V-V_i}{R_i}+\sum_k \frac{V}{R_k}=\sum_j I_j$$ (9)

  Solving for $V$, we get

  $$V=\frac{\sum_i V_i/R_i+\sum_j I_j}{\sum_i 1/R_i+\sum_k 1/R_k}=\frac{\sum I}{\sum G}$$ (10)

  where the reciprocal of the resistance $G=1/R$ is the conductance.

  

  The dual form of the Millman's theorem can be derived based on the loop circuit on the right. Applying KVL to the loop, we have:

  $$\sum_j (I-I_j)R_j+\sum_k I R_k=\sum_i V_i$$ (11)

  Solving for $I$, we get

  $$I=\frac{\sum_i V_i+\sum_j R_jI_j}{\sum_i R_j+\sum_k R_k}=\frac{\sum V}{\sum R}$$ (12)

Example 2: Solve the following circuit:

• Loop current method: Let the three loop currents in the example above be $I_a$, $I_b$ and $I_c$ for loops 1 (top-left bacb), 2 (top-right adca), and 3 (bottom bcdb), respectively, and applying KVL to the three loops, we get

  $$\begin{array}{l} \mbox{loop bacb:}\;\;\;\;\;-V_{s1}+R_1I_a+R_... ...\;-V_{s4}+R_4(I_c-I_a)+R_5(I_c-I_b)+R_6I_c-V_{s6}=0 \end{array}$$ (13)

  We can then solve these 3 loop equations to find the 3 loop currents.

• Node voltage method: If node d is chosen as ground, we can apply KCL to the remaining 3 nodes at a, b, and c, and get (assuming all currents leave each node):

  $$\begin{array}{l} \mbox{node a:}\;\;\;\;\;(V_a-(V_b+V_{s1}))/R... ...;\;\;(V_c-(V_b+V_{s4}))/R_4+(V_c-V_a)/R_3+V_c/R_5=0 \end{array}$$ (14)

  We can then solve these 3 node equations to find the 3 node voltages.

We see that either of the loop-current and node-voltage methods requires to solve a linear system of 3 equations with 3 unknowns.

Example 3: Solve the following circuit with $I=0.5 A$, $V=6 V$, $R_1=3\Omega$, $R_2=8\Omega$, $R_3=6\Omega$, $R_4=4\Omega$. This circuit has 3 independent loops and 3 independent nodes.

• Loop current method:

  Assume three loop currents $I_1$ (left), $I_2$ (right), $I_3$ (top) all in clock-wise direction. We take advantage of the fact that the current source is in loop 1 only, with loop current $I_1=I=0.5A$, and get the following two (instead of three) loop equations with 2 unknown loop currents $I_2$ and $I_3$:

  $$\left\{ \begin{array}{l} R_2(I_2-I_1)+R_3(I_2-I_3)-V=0 \\ R_1(I_3-I_1)+R_4I_3+R_3(I_3-I_2)=0 \end{array} \right.$$ (15)

  i.e.,

  $$\left\{ \begin{array}{l} 14 I_2-6 I_3=10 \\ -6 I_2+13 I_3=1.5 \en... ... \;\;\;\;\; \left\{ \begin{array}{l} I_2=0.952 \\ I_3=0.555 \end{array} \right.$$ (16)

  We can also get the three node voltages with respect to the bottom node treated as ground:
  • Right node: $V_3=-6V$
  • Middle node: $V_2=R_2(I_1-I_2)=8(0.5-0.952)=-3.616V$
  • Left node: $V_1=V_3+R_4 I_3=-3.78 V$

• Node voltage method:

  Assume the three node voltages with respect to the bottom node treated as ground to be $V_1$ (left), $V_2$ (middle), $V_3$ (right). We take advantage of the fact that one side of the voltage source is treated as ground, and get the node voltage $V_3=-V=-6V$. Then we have only two (instead of three) node equations with 2 unknown node voltages $V_1$ and $V_2$:

  $$\left\{ \begin{array}{l} (V_1-V_3)/R_4+(V_1-V_2)/R_1=I \\ (V_2-V_1)/R_1+(V_2-V_3)/R_3+V_2/R_2=0 \end{array} \right.$$ (17)

  i.e.,

  $$\left\{ \begin{array}{l} 7V_1-4V_2=-12 \\ -8V_1+15V_2=-24 \end{a... ...\;\;\;\;\; \left\{ \begin{array}{l} V_1=-3.78 \\ V_2=-3.616 \end{array} \right.$$ (18)

  The voltages are the same as before.

In summary, here we have taken advantage of either the given current source by the following:

• let a given current source be in a loop not shared with any other loop, so that the loop current is known;
• let one of the two ends of a given voltage source be the ground, so that the voltage at the other end is known.

By either of these methods, the number of unknown loop currents or node voltages is reduced by one.

In other words, to simplify the analysis, it is preferable to

• choose independent loops to avoid current source shared by two or more loops,
• choose ground node so that one or more voltage sources are connected to ground.

Example 4: In the circuit below, $V_1=12 V$, $V_2=6 V$, $R_1=3\Omega$, $R_2=8\Omega$, $R_3=6\Omega$, $R_4=4\Omega$.

Find all node voltages with respect to the top-left corner treated as the ground. Then do the same when the middle node where all three resistors $R_1$, $R_2$, and $R_3$ join is treated as ground is treated as the ground.

Answer

Example 5: The two circuits shown below are equivalent, but you may want to choose wisely in terms of which is easier to analyze. Solve this circuit using both node voltage and loop current methods. Assume $R_1=100\Omega$, $R_2=5\Omega$, $R_3=200\Omega$, $R_4=50\Omega$, $V=50V$, and $I=0.2A$.

Answer