Opening Kirchoff!
Kirchoff’s Laws apply to branched and multi-loop circuits. There are two of them, the first states
that the sum of the currents coming into a junction equals the sum of the currents leaving the
junction (charge conservation, no build-up of charge at a junction.
The second states that the sum of the voltage differences around any closed loop is equal to
zero (potential has one value at any given point).
Consider the following circuit with a 10V source and all resistors = 1kΩ. We wish to
determine the current through each resistor and the voltage at points a , b and c. This circuit
can be analyzed either by series and parallel combinations or by Kirchoff’s Laws.
We will use Kirchoff’s Laws and then
check it with the series and parallel
combinations.
Current splits:
I1 = I2+I3+I4
I5 = I3+I4
I5+I2 = I1
Ioops: (clockwise)
L1 : V - I1R1 – I2R2 = 0
L2: I2R2 – I3R3 – I5R5 = 0 I2 = I3+I5 = 1.5 I5 using the line below.
L3: I3R3 – I4R4 = 0
I3 = I4 = ½ I5 using I5 = I3+I4
Using I1 = I5+I2, = I5 + 1.5 I5 or I1 = 2.5 I5
rewrite all currents in terms of I5:
I1 = 2.5 I5
I2 = 1.5 I5
I3 = I4 = ½ I5
L1 : V - I1R1 – I2R2 = 0 = V -1kΩ(2.5 I5 + 1.5 I5) = V – 4 (I5) (1kΩ) = 0
From L1 with V = 10V, 10V - (I5) (4kΩ) = 0 I5 = 2.5 mA
Using the current relations, this gives:
I1 = 2.5 I5
= 6.25mA
I2 = 1.5 I5
= 3.75mA
I3 = I4 = ½ I5 = 1.25mA
Voltages: c = 0V, no resistance between c and battery.
a = 10V – I1R1 = 10v – (6.25mA)(1kΩ.) = 3.75V
b = 0V + I5R5 = (2.5mA)( 1kΩ.) = 2.5V
Check the results using series and parallel combinations to find total resistance and current
from battery I1.
R3 and R4 are in parallel.
The resulting R is in series with R5.
This combination is in parallel with R2.
All of that is in series with R1.
Calculate the result using 1kΩ
resistors:
R3 ll R4 : 1/1kΩ + 1/1kΩ = 1/R = 2/1kΩ or R = 0.5 kΩ
R + R5 = 1.5 kΩ
R2 ll 1.5 kΩ = 1/1kΩ + 1/1.5kΩ = 5 / 3kΩ so R = 0.6kΩ
R1 + 0.6kΩ = 1.6kΩ = Rtot
10V = Itot Rtot = Itot (1.6kΩ)
Itot = 10V / Rtot = 10V / 1.6kΩ = 6.25mA same as before.
The rest follows from the current splits and the equivalent resistances.
After R1, the current splits between 2 branches with R2 = 1kΩ and the combination R = 1.5 kΩ
Since the branches have the same voltage difference across them, the IR products must be
equal. So
I2 = (1.5)I5 and I2+I5 = I1 = 6.25mA = (2.5)I5 giving
I5 = 2.5mA and I2 = 3.75mA
Finally,
I3 = I4 = ½ I5 = 1.25mA
All of which agrees with the previous method. The voltages are calculated using Ohm’s law as
before.
Some circuits cannot be fully analyzed using
series and parallel formulas. For these,
Kirchoff’s Laws are the tools to use. Consider
the following circuit. Unless there are equivalent
resistors in the split diamond configuration, series
and parallel formulas cannot be used to break
down the combination to equivalent resistances.
The structure of the circuit is more complex than
just series and parallel.
For this example, let V = 10 V and the resistors Rx = x kΩ so they are all different. The
directions for the currents are chosen arbitrarily; if they are reversed, the algebra will let us know
with a negative sign. The copy of the diagram below has the currents and loops labeled.
The current rule tells us that
a. I1 = I2 + I3
b. I2 = I4 + I5
c. I4 + I3 = I6
I5 + I6 = I1 = total current.
The loop rule sums ∆V’s around
a loop. Ohm’s law gives us
∆V=IR for a resistor. This is
negative (a drop in V) in the
direction of the current and
positive when going opposite
the current (up hill).
For loop L1, this gives V – I1R1 – I2R2 – I5R5 = 0
L2: +I5R5 – I4R4 –I6R6 = 0
L3: - I2R2 – I5R5 + I6R6 + I3R3 = 0
a. I1 = I2 + I3
L1: V – I1R1 – I2R2 – I5R5 = 0
b. I2 = I4 + I5
L2: +I5R5 – I4R4 –I6R6 = 0
c. I4 + I3 = I6
L3: - I2R2 – I5R5 + I6R6 + I3R3 = 0
I5 + I6 = I1 = total current.
First eliminate some of the currents:
I3 = I1 – I2
I6 = I1 – I5
I4 = I2 – I5
Now substitute these into the equations for loops L2 and L3:
L2: +I5R5 – (I2 – I5)R4 –(I1 – I5)R6 = 0
L3: - I2R2 – I5R5 + (I1 – I5)R6 + (I1 – I2)R3 = 0
Rewrite,
L2: +I5R5 – I2R4 + I5R4 –I1R6 + I5R6 = 0
I5(R5+R4+R6) –I2R4 – I1R6 = 0
I5(R5+R4+R6) = I2R4 + I1R6
or I5 = (I2R4 + I1R6) / (R5+R4+R6)
L3: - I2R2 – I5R5 + I1R6 – I5R6 + I1R3 – I2R3 = 0
- I5(R5+R6) – I2(R2+R3) + I1(R3+R6) = 0
or [- (R5+R6) / (R5+R4+R6)] (I2R4 + I1R6) – I2(R2+R3) + I1(R3+R6) = 0
I1 [(R3+R6) –R6(R5+R6) / (R5+R4+R6)] = I2 [(R2+R3) + R4(R5+R6) / (R5+R4+R6)]
I1 [(R3R5+R3R4+R3R6+R4R6) / (R5+R4+R6)] =
I2[{(R2+R3)(R5+R4+R6) + R4(R5+R6)} / (R5+R4+R6)]
I1 [(R3R5+R3R4+R3R6+R4R6)] = I2 [(R2+R3)(R5+R4+R6) + R4(R5+R6)]
I2 = I1 {[(R3R5+R3R4+R3R6+R4R6)] / [(R2+R3)(R5+R4+R6) + R4(R5+R6)]}
= I1 { } for short.
Return to I5: I5 = (I2R4 + I1R6) / (R5+R4+R6)
I5 = (I1{ }R4 + I1R6) / (R5+R4+R6)
I5 = I1 (R4{ }+R6) / (R5+R4+R6)
L1: V – I1R1 – I2R2 – I5R5 = 0
V = I1R1 + I2R2 + I5R5
V = I1R1 + I1 { }R2 + I1 R5(R4{ }+R6) / (R5+R4+R6)
= I1 [R1 + { }R2 + R5(R4{ }+R6) / (R5+R4+R6) ]
So, with { } = {[(R3R5+R3R4+R3R6+R4R6)] / [(R2+R3)(R5+R4+R6) + R4(R5+R6)]}
I1 = V / [R1 + { }R2 + ( R5(R4{ }+R6) / (R5+R4+R6) ) ]
I5 = I1 (R4{ }+R6) / (R5+R4+R6)
I2 = I1 { }
I3 = I1 – I2
I6 = I1 – I5
I4 = I2 – I5
From these, Ohm’s Law ΔV = IR can be used to calculate voltages.
With numbers for all the resistance values, the expressions all boil down to numbers and don’t
look so terrifying.
Try calculating the currents and voltages at all junctions using
[1] R1 = 0, R2=R3=2kΩ, R4=100Ω, R5=R6=1kΩ
(What is I4 and what are the voltages at b and c?)
[2] R1=0, R2=R6=2kΩ, R4=100Ω, R5=R3=1kΩ