Kirchhoff’s Rules
Dr Miguel Cavero
August 14, 2014
Kirchhoff’s Rules
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Reducing Simple Series/Parallel Circuits
Simple circuits can be analysed using Ohm’s Law and the rules for
series or parallel combinations of resistors.
An equivalent resistance is found, from which the current through, and
voltage across, any resistor can be found.
Consider the following simple circuit:
Find the equivalent resistance, the current through each resistor and
the voltage across each resistor.
Kirchhoff’s Rules
Simple Circuits
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Simple Circuit Example
Find the equivalent resistance.
The parallel combination reduces to (6 × 3)/(6 + 3) = 2 Ω.
Kirchhoff’s Rules
Simple Circuits
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Simple Circuit Example
Find the equivalent resistance.
The equivalent resistance is 4 + 2 = 6 Ω.
Kirchhoff’s Rules
Simple Circuits
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Simple Circuit Example
Find the total current.
Using Ohm’s Law, I = 18/6 = 3 A. In which direction does the current
flow? What is the voltage across the equivalent resistor?
Kirchhoff’s Rules
Simple Circuits
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Simple Circuit Example
What is the current through the 4 Ω resistor?
The total current 3 A must flow through the 4 Ω resistor, and through
the parallel combination. What about the voltage?
Kirchhoff’s Rules
Simple Circuits
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Simple Circuit Example
What is the current through the 6 Ω and 3 Ω resistors?
The total current 3 A is split between the two resistors (current-splitting
rule). The voltage across each must be the same.
Kirchhoff’s Rules
Simple Circuits
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Not So-Simple Circuits
Not all circuits may be reduced to an emf source and an equivalent
resistance.
Consider this circuit:
This circuit cannot be simplified as before.
To analyse and find the currents in this circuit, the techniques
developed by the German physicist Gustav Kirchhoff must be used.
Kirchhoff’s Rules
Kirchhoff’s Rules
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Kirchhoff’s Current Rule
Kirchhoff’s Current Rule (KI):
The sum of the currents entering any junction (or node) must be equal
to the sum of the currents leaving that junction.
KI is often referred to as the junction rule.
This statement is a conservation of charge conservation. Charge
doesn’t build up nor does it disappear at any point.
Kirchhoff’s Rules
Kirchhoff’s Rules
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Kirchhoff’s Voltage Rule
Kirchhoff’s Voltage Rule (KII):
The sum of the potential differences across all components around any
closed circuit loop must be zero.
KII is often reffered to as the loop rule.
The loop rule is a consequence of energy conservation. Any charge
must gain the same amount of energy (from emf sources) as the
energy it loses (through load components or through flowing
”backwards” through an emf source) as it moves through a closed loop.
Kirchhoff’s Rules
Kirchhoff’s Rules
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Electric Potential Rises And Drops
Note: The conventional current direction is used, meaning that current
is taken as the flow of positive charges.
An emf source provides a charge with energy to move from a point of
lower potential to a point of higher potential.
Consider a charge moving from point a to point b, through an emf
source E:
Note the convention that the positive terminal/side of the emf source is
taken as the point of higher potential, and denoted as positive (+).
For a charge moving from point a to point b across the emf source, the
potential increases by a value of ∆Vab = +E.
Kirchhoff’s Rules
Kirchhoff’s Rules
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Electric Potential Rises And Drops
Consider a charge moving from point a to point b, through an emf
source E:
For this charge, the potential decreases by a value of ∆Vab = −E.
The potential decreases from a to b by an amount E (and likewise, it
increases from b to a by an amount E).
Kirchhoff’s Rules
Kirchhoff’s Rules
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Electric Potential Rises And Drops
For a charge moving through a resistor, the potential must drop since
the charge loses energy due to collisions.
Consider a charge moving from point a to point b, through a resistor R:
For this charge, the potential decreases by a value of ∆Vab = −IR.
The potential difference ∆Vab is given by Ohm’s Law and the potential
drops, where I is the current through R.
The potential drops from a to b (in the direction of the current I).
Kirchhoff’s Rules
Kirchhoff’s Rules
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Electric Potential Rises And Drops
Consider a charge moving from point b to point a, through a resistor R:
For this charge, the potential decreases by a value of ∆Vba = −IR.
The potential difference ∆Vab is therefore +IR, and so it increases
from a to b (in the opposite direction of the current I).
Kirchhoff’s Rules
Kirchhoff’s Rules
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Summary
The potential always drops in the direction of the current.
From point a to b, the potential drops on the left-hand side above and
rises on the right-hand side above.
For an emf source, the potential increases or decreases depending on
the orientation of the positive and negative terminals.
Kirchhoff’s Rules
Kirchhoff’s Rules
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Applying Kirchhoff’s Rules
Take the following steps when analysing a circuit using Kirchhoff’s
Rules:
1 Label the circuit (i.e. the junctions/nodes).
2 Label and assign directions to the unknown currents in the circuit.
3 Choose a direction for traversing a loop, and remain consistent in
that direction.
4 Write down the necessary number of Kirchhoff equations (one for
each unknown) and solve simultaneously. There should be at
least one KI equation and at least one KII equation.
Kirchhoff’s Rules
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An Example
Consider the circuit from before:
Label the junctions in the circuit, and label the unknown currents and
assign their directions.
Kirchhoff’s Rules
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An Example
KI (the junction rule) needs to be specified at a junction. It must involve
as many currents as possible.
KI at b: I1 = I2 + I3
Point (junction) e could also have been used, leading to I2 + I3 = I1 .
Note: Since three equations are needed for three unknown currents,
only one of these can be used (since both equations are equal).
Kirchhoff’s Rules
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An Example
KII (the loop rule) needs to be specified for a particular loop and in a
particular direction.
Consider the loop bcdeb - the loop and the direction (clockwise) have
been specified.
Kirchhoff’s Rules
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An Example
For the segment bc, there are no components so the potential remains
the same from b to c.
Consider the segment cd. The current I3 flows from c to d.
The potential therefore drops by an amount Vcd = Vd − Vc = −I3 R3 .
For the segment de, there are no components so the potential once
again remains the same from d to e.
Kirchhoff’s Rules
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An Example
Consider the final segment in the loop, eb. The negative terminal of E2
is connected to point e, while the positive terminal is on the side of
point b.
The potential therefore increases from point e towards point b, by an
amount +E2 .
The current I2 moves from point b towards the emf source E2 . The
potential then drops from point b towards point e.
Kirchhoff’s Rules
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An Example
If the potential drops from b towards e, then the potential must
increase through R2 towards point b, by an amount +I2 R2 .
For the segment eb, the total potential difference is
+E2 + I2 R2
Kirchhoff’s Rules
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An Example
For the loop bcdeb, the KII equation (the sum of all potential
differences around a loop in a given direction is zero) is
KII for bcdeb:
− I3 R3 + E2 + I2 R2 = 0
One more equation is needed to solve for the three unknown currents.
Kirchhoff’s Rules
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An Example
Consider the loop abefa.
The segment be gives
−I2 R2 − E2
The segment fa gives
+E1 − I1 R1
KII for abefa:
Kirchhoff’s Rules
− I2 R2 − E2 + E1 − I1 R1 = 0
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An Example
The three equations are
I1 = I2 + I3
− I3 R3 + E2 + I2 R2 = 0
I2 R2 − E2 + E1 − I1 R1 = 0
Given values for R1 , R2 , R3 , E1 and E2 , these equations are solved
simultaneously to find I1 , I2 and I3 .
Kirchhoff’s Rules
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Example 1
What is the KI equation at point b?
KI at b: I1 = I2 + I3
Kirchhoff’s Rules
Test Yourself
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Example 1
What is the KII equation for the loop bcdeb?
KII for bcdeb:
Kirchhoff’s Rules
− I2 R3 − E2 + E3 + I3 R4 = 0
Test Yourself
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Example 1
What is the KII equation for the loop abefa?
KII for abefa:
Kirchhoff’s Rules
− I1 R1 − I3 R4 − E3 − I1 R2 + E1 = 0
Test Yourself
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Example 1
The three Kirchhoff equations are
KI at b: I1 = I2 + I3
KII for bcdeb:
KII for abefa:
Kirchhoff’s Rules
− I2 R3 − E2 + E3 + I3 R4 = 0
− I1 R1 − I3 R4 − E3 − I1 R2 + E1 = 0
Test Yourself
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Example 1
The previous three Kirchhoff equations suffice. However, one more KII
is possible for the loop abcdefa.
−I1 R1 − I2 R3 − E2 − I1 R2 + E1 = 0
Kirchhoff’s Rules
Test Yourself
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Example 2
Write down three KI equations and four KII equations, which would be
used to solve for the unknown currents.
Kirchhoff’s Rules
Test Yourself
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Example 2
KI at A:
I1 = I3 + I4
KI at B: I4 = I5 + I6
KI at D:
Kirchhoff’s Rules
I2 + I6 = I1
Test Yourself
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Example 2
KII ACBA:
− I3 R3 − E3 + I5 R5 + I4 R4 = 0
KII ABDA:
− I4 R4 − I6 R6 + E1 − I1 R1 = 0
KII BCDB:
− I5 R5 − I2 R2 + E2 + I6 R6 = 0
KII ACDA:
Kirchhoff’s Rules
− I3 R3 − E3 − I2 R2 + E2 + E1 − I1 R1 = 0
Test Yourself
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Example 1 Revisited - Solving Simultaneously
Let the resistors and emf sources have the following values:
R1 = 6 Ω,
R2 = 4 Ω,
E1 = 19 V,
R3 = 4 Ω
R4 = 1 Ω
E2 = 6 V E3 = 2 V
Use Kirchhoff’s Rules to solve for the unknown currents.
Kirchhoff’s Rules
Solving For Unknown Currents
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Example 1 Revisited - Solving Simultaneously
The three Kirchhoff equations are
KI at b: I1 = I2 + I3
KII for bcdeb:
KII for abefa:
Kirchhoff’s Rules
− I2 R3 − E2 + E3 + I3 R4 = 0
− I1 R1 − I3 R4 − E3 − I1 R2 + E1 = 0
Solving For Unknown Currents
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Example 1 Revisited - Solving Simultaneously
Substituting the values for the resistors and emf sources gives:
KI at b: I1 = I2 + I3
KII for bcdeb:
KII for abefa:
Kirchhoff’s Rules
− 4I2 − 6 + 2 + I3 = 0
− 6I1 − I3 − 2 − 4I1 + 19 = 0
Solving For Unknown Currents
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Example 1 Revisited - Solving Simultaneously
KI at b : I1 = I2 + I3
(1)
KII for bcdeb : −4I2 − 6 + 2 + I3 = 0
(2)
KII for abefa : −6I1 − I3 − 2 − 4I1 + 19 = 0
(3)
Substitute Equation (1) into Equation (3).
KII for abefa : −6(I2 + I3 ) − I3 − 2 − 4(I2 + I3 ) + 19 = 0
KII for abefa : −10I2 − 11I3 + 17 = 0
(4)
Multiple Equation (2) by 11, and add it to Equation (5).
KII for bcdeb : −44I2 + 11I3 − 44 = 0
(5)
KII for abefa : −10I2 − 11I3 + 17 = 0
∴
Kirchhoff’s Rules
: −54I2 − 27 = 0
(6)
I2 = −0.5 A
(7)
Solving For Unknown Currents
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Example 1 Revisited - Solving Simultaneously
The significance of the negative sign in (7) (I2 = −0.5 A) is that the
current flows in the opposite direction than what was initially chosen I2 flows from point e to d, to c and then to b.
Substitute (7) into Equation (2)
I2 = −0.5 A
KII for bcdeb : −4(−0.5) − 6 + 2 + I3 = 0
Kirchhoff’s Rules
∴
I3 = 4 − 2 = 2 A
(8)
∴
I1 = −0.5 + 2 = 1.5 A
(9)
Solving For Unknown Currents
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Example 1 Revisited - Given An Initial Value
Let the resistors and emf sources have the following values:
R1 = 6 Ω,
R2 = 4 Ω,
E1 = 19 V,
R3 = 4 Ω
R4 = 1 Ω
E2 = 6 V E3 = 2 V
Use Kirchhoff’s Rules to solve for the unknown currents.
Kirchhoff’s Rules
Solving For Unknown Currents
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Example 1 Revisited - Given An Initial Value
The three Kirchhoff equations are
KI at b: I1 = I2 + I3
KII for bcdeb:
KII for abefa:
Kirchhoff’s Rules
− 4I2 − 6 + 2 + I3 = 0
− 6I1 − I3 − 2 − 4I1 + 19 = 0
Solving For Unknown Currents
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Example 1 Revisited - Given An Initial Value
Given that I2 = −0, 5 A, find I1 and I3 .
KI at b : I1 = I2 + I3
(10)
KII for bcdeb : −4I2 − 6 + 2 + I3 = 0
(11)
KII for abefa : −6I1 − I3 − 2 − 4I1 + 19 = 0
(12)
KII for bcdeb : −4(−0.5) − 6 + 2 + I3 = 0
Kirchhoff’s Rules
∴
I3 = 4 − 2 = 2 A
(13)
∴
I1 = −0.5 + 2 = 1.5 A
(14)
Solving For Unknown Currents
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