Many circuits are too complex to be solved using the rules for series or parallel circuits or the techniques for conversion to simpler circuits described in previous chapters. For these circuits we need more general solution methods. The most general method is given by Kirchhoff’s laws, which permit the calculation of all circuit voltages and currents of circuits by a solution of a system of linear equations.
There are two Kirchhoff laws, the voltage law and the current law. These two laws can be used to determine all voltages and currents of circuits.
Kirchhoff’s voltage law (KVL) states that the algebraic sum of the voltage rises and voltage drops around a loop must be zero.
A loop in the above definition means a closed path in the circuit; that is, a path that leaves a node in one direction and returns to that same node from another direction.
In our examples, we will use clockwise direction for loops; however, the same results will be obtained if the counterclockwise direction is used.
In order to apply KVL without error, we have to define the so called reference direction. The reference direction of the unknown voltages points from the + to the – sign of the assumed voltages. Imagine using a voltmeter. You would place the voltmeter positive probe (usually red) at the component’s reference + terminal. If the real voltage is positive, it is in the same direction as we assumed, and both our solution and the voltmeter will show a positive value.
When deriving the algebraic sum of the voltages, we must assign a plus sign to those voltages where the reference direction agrees with the direction of the loop, and negative signs in the opposite case.
Another way to state Kirchhoff’s voltage law is: the applied voltage of a series circuit equals the sum of the voltage drops across the series elements.
The following short example shows the use of Kirchhoff’s voltage law.
Find the voltage across resistor R_{2, } given that the source voltage, V_{S} = 100 V and that the voltage across resistor R_{1 }is V_{1} = 40 V.
The figure below can be created with TINA Pro Version 6 and above, in which drawing tools are available in the schematic editor.
The solution using Kirchhoff’s voltage law: V_{S} + V_{1} + V_{2 }=0, or V_{S} = V_{1} + V_{2}
hence: V_{2} = V_{S} – V_{1} = 10040 = 60V
Note that normally we don’t know the voltages of the resistors (unless we measure them), and we need to use both Kirchhoff’s laws for the solution.
Kirchhoff’s current law (KCL) states that the algebraic sum of all the currents entering and leaving any node in a circuit is zero.
In the following, we give a + sign to currents leaving a node and a  sign to currents entering a node.
Here’s a basic example demonstrating Kirchhoff’s current law.
Find the current I_{2} if the source current I_{S} = 12 A, and
I_{1} = 8 A.
Using Kirchhoff’s current law at the circled node: I_{S} + I_{1} + I_{2} = 0 , hence:
I_{2}= I_{S} – I_{1} = 12 – 8 = 4 A,
as you can check using TINA (next
figure).
In the next example, we will use both Kirchhoff’s laws plus Ohm’s law to calculate the current and the voltage across the resistors.
In the figure below, you will note the Voltage Arrow above resistors. This is a new component available in Version 6 of TINA and works like a voltmeter. If you connect it across a component, the arrow determines the reference direction (to compare to a voltmeter, imagine placing the red probe at the tail of the arrow and the black probe at the tip). When you run DC analysis, the actual voltage on the component will be displayed on the arrow.
According to Kirchhoff’s voltage law: V_{S} = V_{1}+V_{2}+V_{3}
Now using Ohm’s law: V_{S}=I*R_{1}+ I*R_{2}+I *R_{3}
And from here the current of the circuit:
I=V_{S }/(R_{1}+R_{2}+R_{3})= 120/(10+20+30) = 2 A
Finally the voltages of the resistors:
V_{1}=I*R_{1} = 2*10 = 20 V; V_{2} = I*R_{2} = 2*20 = 40 V; V_{3} = I*R_{3} =2*30 = 60 V
The same results will be seen on the Voltage Arrows by simply running TINA’s interactive DC analysis.
The total number of independent applications of Kirchhoff’s laws in a circuit is the number of circuit branches, while the total number of unknowns (the current and voltage of each branch) is twice that . However, by also using Ohm’s law at each resistor and the simple equations defining the applied voltages and currents, we get a system of equation where the number of unknowns is the same as the number of equations.
Find the branch currents I1, I2, I3 in the circuit below.
The nodal equation for the circled node:

I_{1 } I_{2
} I_{3} = 0
or multiplying by 1
I_{1 }+ I_{2 }+ I_{3} = 0
The loop equations (using the clockwise direction) for the loop L1, containing V_{1}, R_{1} and R_{3 }
V_{1}+I_{1}*R_{1}I_{3}*R_{3} = 0
and for the loop L2, containing V_{2}, R_{2} and R_{3}
I_{3}*R_{3} – I_{2}*R_{2} +V_{2} = 0
Substituting the component values:
I_{1}+ I_{2}+ I_{3} = 0 8+40*I_{1} – 40*I_{3} = 0 40*I_{3} –20*I_{2} +16 = 0
Express I_{1} using the nodal equation: I_{1}
= I_{2} – I_{3}
then substitute it into the second equation:
V_{1 }– (I_{2} + I_{3})*R_{1} –I_{3}*R_{3 } = 0 or –8 (I_{2} + I_{3})*40 – I_{3}*40 = 0
Express I_{2} and substitute it into the third equation, from which you can already calculate I_{3}:
I_{2} =  (V_{1} + I_{3}*(R_{1}+R_{3}))/R_{1} or I_{2} = (8 + I_{3}*80)/40
I_{3}*R_{3} + R_{2}*(V_{1} + I_{3}*(R_{1}+R_{3}))/R_{1} +V_{2} = 0 or I_{3}*40 + 20*(8 + I_{3}*80)/40 + 16 = 0
And: I_{3} =  (V_{2} + V_{1}*R_{2}/R_{1})/(R_{3}+(R_{1}+R_{3})*R_{2}/R_{1}) or I_{3} = (16+8*20/40)/(40 + 80*20/40)
Therefore I_{3} =  0.25 A; I_{2} = (80.25*80)/40 = 0.3 A and I_{1} =  (0.30.25) =  0.05 A
Or: I_{1} = 50 mA; I_{2} = 300 mA; I_{3} = 250 mA.
Now let’s solve the same equations with TINA’s
interpreter:
{Solution by TINA's Interpreter}
Sys I1,I2,I3 I1+I2+I3=0 V1+I1*R1I3*R3=0 I3*R3I2*R2+V2=0 end; I1=[50m] I2=[300m] I3=[250m] 
Finally let’s check the results using TINA:
Next, let’s analyze the following even more complex circuit and determine its
branch currents and voltages.

I_{L }+ I_{R1 } I_{s }= 0_{ }(for N1) _{ }
 I_{R1 }+ I_{R2 }+ I_{s3} = 0 (for N2)
V_{s1 } V_{R3 }+ V_{Is }+ V_{L} = 0 (for L1)
V_{Is }+ V_{s2} +V_{R2 }+V_{R1} = 0 (for L2)
V_{R2 } V_{s2 }+ V_{s3} = 0 (for L3)
Applying Ohm’s law:
V_{L }= I_{L}*R_{L }
_{ }V_{R1 }=I_{R1}*R_{1}
V_{R2} = I_{R2}*R_{2}
V_{R3} =  I_{L}*R_{3}
This is 9 unknowns and 9 equations. The easiest way to solve this is to use TINA’s
interpreter. However, if we are pressed to use hand calculations, we note that this set of equations can be easily reduced to a system of 5 unknowns by substituting the last 4 equations into the L1, L2, L3 loop equations. Also, by adding equations (L1) and (L2), we can eliminate V_{Is }, reducing the problem to a system of 4 equations for 4 unknowns (I_{L, }I_{R1 } I_{R2, }I_{s3}). When we have found these currents, we can easily determine V_{L ,} V_{R1}, V_{R2, } and V_{R3} using the last four equations (Ohm’s law).
Substituting V_{L },V_{R1,}V_{R2} ,V_{R3} :
I_{L }+ I_{R1 } I_{s }= 0_{ }(for N1) _{ }
 I_{R1 }+ I_{R2 }+ I_{s3} = 0 (for N2)
V_{s1 }+ I_{L}*R_{3 }+ V_{Is }+ I_{L}*R_{L} = 0 (for L1)
V_{Is }+ V_{s2} + I_{R2}*R_{2 }+ I_{R1}*R_{1} = 0 (for L2)
 I_{R2}*R_{2 } V_{s2 }+ V_{s3} = 0 (for L3)
Adding (L1) and (L2) we get
I_{L }+ I_{R1 } I_{s }= 0_{ }(for N1) _{ }
 I_{R1 }+ I_{R2 }+ I_{s3} = 0 (for N2)
V_{s1 }+ I_{L}*R_{3 }+ I_{L}*R_{L} + V_{s2} + I_{R2}*R_{2 }+ I_{R1}*R_{1} = 0 (L1) + (L2)
 I_{R2}*R_{2 } V_{s2 }+ V_{s3} = 0 (for L3)
After substituting the component values, the solution to these equations comes readily.
I_{L}+I_{R1 }– 2 = 0 (for N1)
I_{R1} + I_{R2} + I_{S3} = 0 (for N2)
120  + I_{L}*90 + I_{L}*20 + 60 + I_{R2}*40 + I_{R1}*30 = 0 (L_{1}) + (L_{2})
I_{R2}*40 – 60 + 270 = 0 (for L_{3})
from L_{3} I_{R2} = 210/40 = 5.25 A (I)
from N_{2} I_{S3} – I_{R1} =  5.25 (II)
from L_{1}+L_{2} 110 I_{L} + 30 I_{R1} = 150 (III)
and for N_{1} I_{R1} – I_{L} = 2 (IV)
Multiply (IV) by –30 and add to (III) 140 I_{L} = 210 hence I_{L} =  1.5 A
Substitute I_{L} into (IV) I_{R1} = 2 + (1.5) = 0.5 A
and I_{R1} into (II) I_{S3} = 5.25 + I_{R1} = 4,75 A
And the voltages: V_{R1} = I_{R1}*R_{1} = 15 V; V_{R2} = I_{R2}*R_{2} = 210 V;
V_{R3} =  I_{L}*R_{3}= 135 V; V_{L} = I_{L}*R_{L} =  30 V; V_{Is} = V_{S1}+V_{R3}V_{L} = 285 V
{Solution of the original
equations by TINA's Interpreter}

Solution of the reduced set of equations using the interpreter:
{Solution of the reduced set of equations by TINA's Interpreter}
Sys Il,Ir1,Ir2,Is3 Il+Ir12=0 Ir1+Ir2+Is3=0 120+110*Il+60+40*Ir2+30*Ir1=0 40*Ir2+210=0 end; Il=[1.5] Ir1=[500m] Ir2=[5.25] Is3=[4.75] 
We can also enter expressions for the voltages and have TINA’s Interpreter calculate them:
Il:=1.5;
Ir1:=0.5; Ir2:=5.25; Is3:=4.75; Vl:=Il*RL; Vr1:=Ir1*R1 Vr2:=Ir2*R2; Vr3:=Il*R3; VIs:=Vs1Vl+Vr3; Vl=[30] Vr1=[15] Vr2=[210] Vr3=[135] VIs=[285] 