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 source 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.

• The Branch-Current Method:
  • Step 1: Label each unknown branch current arbitrarily with a reference direction. If the calculated value is negative, the actual direction of the current is opposite to the reference direction.
  • Step 2: Define unknown voltages in terms of the assumed currents to obtain three element equations:
    
    $$\begin{array}{l} V_{ab}=2I_1,\;\;\mbox{or}\;\;\;V_{ba}=-2I_1... ...3 \\ V_{cb}=4I_2,\;\;\mbox{or}\;\;\;V_{bc}=-4I_2 \end{array}$$
    
    

  • Step 3: Apply KCL to node b:
    
    $$\sum I_b=0=+I_1+I_2-I_3\;\;\;\;\;\;\;\;\;\;(1)$$
    
    
    We could also apply KCL to node d:
    
    $$\sum I_d=0=-I_1-I_2+I_3$$
    
    
    but this equation is not independent as it adds no information.
  • Step 4: Apply KVL to loops abda and bcdb:
    
    $$\begin{array}{r} \sum V_{abda}=0=V_{ab}+V_{bd}+V_{da}\;\;\;... ...cdb}=0=V_{cb}+V_{bd}+V_{dc}\;\;\;\;\;\;\;\;\;\;(3) \end{array}$$
    
    
    We could also apply KVL to loop abcda:
    
    $$\sum V_{abcda}=0=V_{ab}+V_{bc}+V_{cd}+V_{da}$$
    
    
    But this equation is not independent as it can be obtained as the difference between the previous two equations. The above three independent equations (one for current, two for voltage) are called connection equations.
  • Step 5: We now have three simultaneous equations with three unknown branch currents:
    
    $$\begin{array}{r} \sum I_b=0=+I_1+I_2-I_3\;\;\;\;\;\;\;\;\;\;... ..._{bd}+V_{dc}=4I_2+8I_3-20\;\;\;\;\;\;\;\;\;\;(3') \end{array}$$
    
    
    which can be rewritten as
    
    $$\left\{ \begin{array}{rrrrr} I_1 & +I_2 & -I_3& = & 0\;\;\;\... ..._2& +8I_3 & = & 20\;\;\;\;\;\;\;\;\;\;(3') \end{array} \right.$$
    
    

  • Step 6: Solving the equations, we get the three unknown currents:
    
    $$I_1=4,\;\;\;I_2=-1,\;\;\;I_3=3$$
    
    
    and then we get the three unknown voltages:
    
    $$V_{ab}=8,\;\;\;V_{bd}=24,\;\;\;V_{cb}=-4$$
    
    

  Note that equations from KCL at node d and KVL to other loops are not independent. In general, if a circuit has n nodes and b branches, then there are $(n-1)$ independent node equations and $l=b-(n-1)$ independent loop equations. In other words, the sum of the number of independent loops and the number of independent nodes is always the same as the number of branches, i.e., the number of equations is always equal to the number of unknowns in the branch current method.

• The Loop/Mesh-Current Method (based on KVL):
  1. Define a loop current around each loop in clockwise direction (although it could be arbitrary). Assume there are $l$ independent loops in the circuit, then we have $l$ loop currents as the unknown variables.
  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.

  Now we can resolve the problem above using loop-current method:

  

  For the first circuit, we apply KVL to the two loops to get
  

  $$\sum V_{abda} =0=-32+2I_a+8(I_a-I_b)$$
  
  
  
  
  $$\sum V_{bcdb} =0=8(I_b-I_a)+4I_b+20$$
  
  
  Rewrite these as:
  
  $$\left\{ \begin{array}{rrrr} 10I_a&-8I_b&=&32 -8I_a &+12I_b&=&-20 \end{array} \right.$$
  
  
  which can be solved to get $I_a=I_1=4$, $I_b=-I_2=1$, and $I_3=I_a-I_b=3$.

• The Node-Voltage Method (based on KCL):
  1. Assume there are n nodes in the circuit. Select one node as the ground, i.e., the reference point for all voltages of the circuit. The voltage at each of the remaining n-1 nodes is an unknown to be obtained.
  2. Apply KCL to each of the n-1 nodes to obtain n-1 equations.
  3. Solve the equation system with n-1 equations for the n-1 unknown node voltages.

  Assume node $d$ is the ground, and consider just $V_b=V_{bd}$ as the only unknown in the problem. Apply KCL to node $b$, we have
  

  $$\sum I_b=0=I_1+I_2-I_3$$
  
  
  where
  
  $$I_1=\frac{V_a-V_b}{R_{ab}}=\frac{32-V_b}{2},\;\;\;\; I_2=\fr... ...ac{20-V_b}{4},\;\;\;\; I_3=\frac{V_b}{R_{bd}}=\frac{V_b-0}{8}$$
  
  
  This equation of one unknown $V_b$ can be solved to get $V_b=24$, and all other unknown currents and voltages can be found easily.

  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 $b$, 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:
  

  $$\sum_i \frac{V-V_i}{R_i}-\sum_j I_j+\sum_k \frac{V}{R_k}=0$$
  
  
  Solving for $V$, we get

  
  

  $$V=\frac{\sum_i V_i/R_i+\sum_j I_j}{\sum_i 1/R_i+\sum_k 1/R_k}=\frac{\sum I}{\sum G}$$
  
  
  where the reciprocal of the resistance $G=1/R$ is called the conductance.

  

In summary,

• Branch current method: each equation is for one of the branches.
• Loop current method: each equation is for one of the independent loops.
• Node voltage method: each equation is for one of the independent nodes.

Example 2: Solve the following circuit:

• Branch current method:

  There are $n=4$ nodes (a,b,c,d) and $b=6$ branches in the circuit, therefore we can get $n-1=4-1=3$ independent node equations (first three) and $b-(n-1)=6-3=3$ independent loop equations (second three):
  

  $$\begin{array}{l} \mbox{node a:}\;\;\;\;\; I_1+I_2-I_3=0 \\ ... ...}\;\;\;\;\;-V_{s4}+I_4R_4+I_5R_5-I_6R_6-V_{s6}=0 \end{array}$$
  
  
  Solving these 6 equations we get the 6 branch currents.

• Loop current method: Let the three loop currents in the example above be $I_a$, $I_b$ and $I_c$ for loops 1, 2, and 3, respectively, and applying KVL to the three loops, we get
  
  $$\begin{array}{l} \mbox{loop 1:}\;\;\;\;\;-V_{s1}+R_1I_a+R_3(... ...V_{s4}+R_4(I_c-I_a)+R_5(I_c-I_b)+R_6I_c-V_{s6}=0 \end{array}$$
  
  
  We can then solve these 3 loop equations to find the 3 loop currents.
• Node voltage method: Choose node d as ground, and then apply KCL to the remaining 3 nodes to get:
  
  $$\begin{array}{l} \mbox{node a:}\;\;\;\;\;(V_a-V_b-V_{s1})/R_... ...\;\;(V_c-V_b-V_{s4})/R_4+(V_c-V_a)/R_3+V_c/R_5=0 \end{array}$$
  
  
  We can then solve these 3 node equations to find the 3 node voltages.

Example 3: Solve the following circuit.

$I=0.5 A$, $V=6 V$, $R_1=3\Omega$, $R_2=8\Omega$, $R_3=6\Omega$, $R_4=4\Omega$.

Loop current method:

Assume three loop currents $I_1$ (left), $I_2$ (right), $I_3$ (top). We have

$$I_1=I=0.5A$$

$$\left\{ \begin{array}{l} R_2(I_2-I_1)+R_3(I_2-I_3)-V=0 \\ ... ...l} 14 I_2-6 I_3=10 \\ -6 I_2+13 I_3=1.5 \end{array} \right.$$

Solving to get: $I_2=0.952 A$, $I_3=0.555 A$. We can also get the three node voltages: $V_3=-6V$ (right), $V_2=R_2(I_1-I_2)=8(0.5-0.952)=-3.616V$ (middle), and $V_1=V_3+R_4 I_3=-3.78 V$.

Node voltage method:

Assume the bottom node is ground and the three node voltages are $V_1$ (left), $V_2$ (middle), $V_3=V=-6V$ (right).

$$\left\{ \begin{array}{l} (V_1-V_3)/R_4+(V_1-V_2)/R_1=I \\ ... ...ay}{l} 7V_1-4V_2=-12 \\ -8V_1+15V_2=-24 \end{array} \right.$$

Solving for $V_1$ and $V_2$, we get: $V_1=-3.78$, $V_2=-3.616$, same as before.

Example 4: (Homework) Find all node voltages with respect to the top-left corner treated as reference node:

$V_1=12 V$, $V_2=6 V$, $R_1=3\Omega$, $R_2=8\Omega$, $R_3=6\Omega$, $R_4=4\Omega$.

Answer

Note: While using node voltage and loop current methods to solve a given circuit, to simplify the analysis, it is preferable to

• choose independent loops to avoid current source shared by two loops,
• choose ground node so that the voltage source is connected to ground.

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