Kirchhoff's rules

Three circuit elements have been considered thus far, the emf, the resistor, and the capacitor.  In circuit diagrams they are represented by the symbols shown below.

Most interesting circuits consist of more than just one emf and one resistor.  Some circuits have more than one resistor connected in series, as shown in the figure below.

In such a circuit the same current I flows through each resistor.  The potential difference across resistor 1 is V1 = IR1, the potential difference across resistor 2 is V2 = IR2, and the potential difference across resistor 3 is V3 = IR3.  The potential difference across the chain is equal to the battery voltage V.  We have V = V1+V2+V3 = IR1+IR2+IR3.  The equivalent resistance of the circuit can be calculated from V = IR.  The two equation yield

R = R1 + R2 + R3.

The equivalent resistance of any number of resistors connected in series is the sum of their individual resistances.

In the figure above, let R1 = 4W, R2 = 4W, and R3 = 12W.  Then the equivalent resistance is R = 20W.  If the battery voltage is 10V, then the current in the circuit is I = V/R = (10/20)A = 0.5 A.  The current through each resistor is 0.5 A.  The power dissipated is P = IV = 5W.
Assume another 20W resistor is added to the chain.  Adding another 20W resistor increases the equivalent resistance to R = 40W.  The current drops to I = (10/40)A = 0.25A.  The power dissipated drops to P = 2.5W.
Adding an additional resistance in series
reduces the current in the circuit,
reduces the power I2Ri dissipated by each of the i resistances Ri,
reduces the total power dissipated.

The light bulbs in a chain of Christmas lights are often connected in series.  If 10 identical light bulbs are in the chain and the voltage across the chain is 120V, then the voltage across each bulb is 12V.  The same current flows through each bulb.  If one bulb burns out, no current flows, and none of the bulbs will light up.

It is more common to have more than one resistance connected in parallel, as shown in the figure below.

The current I from the battery is divided into I1, I2, and I3 flowing through R1, R2, and R3, respectively.  These three currents recombine when the branches meet again.

I = I1 + I2 + I3.

The voltage across each resistor in the parallel circuit shown is equal to the battery voltage V.  We have I1 = V/R1, I2 = V/R2, I3 = V/R3.  We find the equivalent resistance from V = IR,

1/R = I/V = (I1 + I2 + I3)/V = (1/R1) + (1/R2) + (1/R3).

1/R = (1/R1) + (1/R2) + (1/R3).

For a set of parallel resistors the reciprocal of their equivalent resistance equals the sum of the reciprocals of their individual resistances.

Assume that in the circuit shown above R1 = 8W, R2 = 8W, and R3 = 4W.  Then 1/R = (1/8W)+(1/8W)+(1/4W) = (1/2W). R=2W.  If the battery voltage is 10V, then the current in the circuit is I = V/R = (10/2) = 5A.  The individual currents are found from Ii = V/Ri.  The voltage across each resistor is 10V, therefore
I1 = 10/8 = 1.25 A,
I2 = 10/8 = 1.25 A ,
I3 = 10/4 = 2.5 A .
The currents add to yield the total current I = 5A.
If another 8W resistor is added in parallel, a current of 1.25A will flow through this resistor.  The total current in the circuit increases, the equivalent resistance decreases, and the power P = IV dissipated by the circuit increases.
Note: The equivalent resistance of any parallel combination of resistors is always less than any of the individual resistances.

If 10 Christmas lights are connected in parallel, then one burned-out bulb does not prevent the other bulbs from lighting up.  However, a different type of bulb must be used if the bulbs are to be connected parallel to a 120V outlet, because the voltage across each bulb will be 120V.

Often circuits contain one emf and a number of series and parallel connections.  An example is the section of a circuit shown below.

The total resistance of such a circuit can be found by sequentially replacing parts of the circuit with a single equivalent resistance.  The equivalent circuit then consists of a single emf and a single resistor.  This procedure yields the total current flowing in the circuit, and the currents flowing through the individual resistors can be found by reversing the reduction process.

Let us find the total resistance between the points A and B in the circuit above.  R1 and R2 are in series and are replaced by R12 = R1 + R2.

Now R12 and R3 are in parallel and can be replaced by R, where 1/R = 1/R1+1/R2.
If R1 = R2 = R3 = 1W, then R12 = 2W and R = (2/3)W.  If a 6V battery is connected across points A and B, then the current flowing in the circuit is I = V/R = 9A.  The current flowing through R3 is I = V/R3, the current flowing through R1 and R2 is V/R12 = 3A.

General rules for finding the equivalent resistance of a simple circuit:

Two (or more) resistors with their heads directly connected together and their tails directly connected together are in parallel.  They can be replaced by one equivalent resistor R using 1/R = (1/R1) + (1/R2) + (1/R3) + ….
Two (or more) resistors connected together so that the tail of one is connected to the head of the next, with no other path for the current to take along the line connecting them, are in series.  They can be replaced by one equivalent resistor R using R = R1 + R2 + R3+….
For resistors in series, the same current flows through each resistor, and for resistors in parallel, the same voltage drops across each resistor.

A junction in a circuit is a point where at least three circuit paths meet.  A branch is a path connecting two junctions.  The circuit below has two junctions, labeled A and B, and three branches, namely the three different paths from A to B.

This circuit is a multi-loop circuit, with more than one battery in different branches of the circuit.  To analyze such a circuit and to find the currents in all branches of the multi-loop circuit one must use Kirchhoff's rules.

Kirchhoff's first rule :
Junction rule:
At any junction point in a circuit where the current can divide, the sum of the currents into the junction must equal the sum of the currents out of the junction.  (This is a consequence of charge conservation.)
Kirchhoff's second rule :
Loop rule:
When any closed circuit loop is traversed, the algebraic sum of the changes in the potential must equal zero.  (This is a consequence of conservation of energy.)

Kirchhoff's rules apply to all circuit.

General procedure for analyzing a circuit using Kirchhoff's:

Replace any combination of resistors in series or parallel with their equivalent resistance.  Label each remaining resistor and each emf with a symbol.
Choose a direction for the current in each branch of the circuit, and label each current.
Add plus and minus signs to label the high and low potential sides of each circuit element, (i.e. each emf, resistor, capacitor).  Label the side of the resistor on which the current enters with a plus sign (+) and the side on which the current exits with a minus sign (-).
Sometimes it is hard to predict in which direction the current will flow in a particular loop.  Simply pick a direction.  If the real current flows in the opposite direction, the current in your solution will be negative.  The minus sign just indicates that the current flows in the direction opposite to one you picked.
Apply Kirchhoff's first rule to all but one of the junctions in the circuit.
Each time you use the junction rule a current that has not been used before must be included.
Apply Kirchhoff's second rule to as many loops as needed to obtain as many equations (including the junction equations) as you have unknowns.
Each time you use the loop equation you have to include a circuit element that has not been used before.
To write down a loop equation, you choose a starting point, and then follow a path around the loop in one direction until you get back to the starting point.  As you cross a battery or a resistors note the change in voltage.  If you cross from - to +, the change is positive.  If you cross from + to -, the change is negative.  Add these changes in voltage and set the sum equal to zero.
Solve the equations to obtain the values of the unknowns.

Example:

In the circuit below, let V1 be a 1V battery,  V2 be a 2V battery, V3 be a 3V battery, and let each resistor be a 1W resistor.  Pick directions for the currents as shown in the diagram.

Image1208a.gif (3330 bytes)

Junction A:  I1+I3 = I2
Loop 1: V1-I1R1-I2R2-V2-I1R4 = 0
This yields
1V-I11W-I21W-2V-I11W = 0,   or   2I1+I2 = -1A.  (Units: V/W = A)
Loop 2: V3-I3R3-I2R2-V2 = 0
This yields
3V-I31W-I21W-2V = 0,  or  I3+I2 = 1A.
We now have to solve the three equations,
I1+I3 = I2,
2I1+I2 = -1A,
I3+I2 = 1A,
for the three unknown currents.
Eliminate I3:  I3 = 1A-I2 using the third equation.  The first two equations the yield
I1+1A-I2 = I2,    I1-2I2 = -1A,
2I1+I2 = -1A.
Eliminate I2:  I2 = -1A-2I1.  We now can solve for I1.
I1+2A+4I1 = -1A,    5I1 = -3A,   I1 = -(3/5)A.
Now we can solve for I2 and I3.
I2 = -1A-2I1 = -1A+(6/5)A = (1/5)A
I3 = I2-I1 = (4/5)A
I2 and I3 flow in the directions picked in the diagram, I1 flows in a direction opposite to the one picked in the diagram.

Voltage sources

Batteries and other voltage sources have internal resistance.  When they do work moving charges against the electric force, some of this work is already converted to thermal energy in the battery.  The amount of energy lost to thermal energy in the battery is Ir, where I is the current flowing in the circuit and r is the internal resistance of the battery.  The voltage across the battery terminals therefore drops from the nominal value V to (V-Ir) when a current is flowing in the circuit.  In a circuit diagram we represent the internal resistance of the battery by a resistor r connected in series with the emf.

Image1203a.gif (1226 bytes)

Electrical Measurements

A voltmeter is a device used to measure voltages, while an ammeter measures currents.  Meters are either analog or digital devices.  Analog meters show the output on a scale with a needle, while digital devices produce a digital readout.  Analog voltmeters and ammeters are both based on a device called a galvanometer.  Digital voltmeters and ammeters generally determine the voltage drop across a known resistor and then convert the result to a digital value for display.

Voltmeters

Resistors in parallel have the same voltage across them.  If you want to measure the voltage across a circuit element, such as a resistor, you place the voltmeter in parallel with the resistor.  The voltmeter is shown in a circuit diagram as a V in a circle, and it acts as another resistor.  To prevent the voltmeter from changing the current in the circuit (and therefore the voltage across the resistor), the voltmeter must have a resistance much larger than that of the resistor.  If the resistance of the voltmeter is large, only a negligible current flows through the meter.

Ammeters

Resistors in series have the same current flowing through them.  An ammeter, must be placed in series with a resistor to measure the current through the resistor.  On a circuit diagram, an ammeter is shown as an A in a circle. The ammeter acts as a resistor.  To prevent the ammeter from changing the current in the circuit, it must have a very small resistance compared to the resistance R of the circuit.

Links:

Circuit Construction Kit
A two-resistor circuit
A four-resistor circuit
The Wheatstone bridge

Problem:

Find the equivalent resistance between the points A and B of the circuit shown in the figure below.

resistance.gif (5397 bytes)

Solution:

You can find the equivalent resistance using Kirchhoff's rules.
The junction rule states that the sum of the currents entering a junction must equal the sum of the currents leaving that junction.  The loop rule states that the sum of the potential differences around any closed circuit loop must be zero.

How to apply Kirchhoff's rules.
Assume some direction for the current flowing in each part of the circuit.  

If a resistor is traversed in the direction of the current, the change in potential across the resistor is –IR.

If a resistor is traversed in the direction opposite to that of the current, the change in potential across the resistor is +IR.

If a voltage source is traversed in the direction from – to +, the change in potential is +V.

If a voltage source is traversed in the direction from + to -, the change in potential is -V.

You can use the junction rule as many times as it is possible to include in it a current that has not been used in a previous junction rule equation.  The number of equations must be equal to the number of unknowns.

Now apply the rules to you circuit.
Assume you connect a battery between A and B so that A is at some voltage V and B is at ground.  A current I will start flowing through the circuit from A to B.  V=IR, R=V/I.  If you know I, then you know R.

To find R for your circuit we need to know the currents flowing through the 6 resistors.  Let I(1) denote the current flowing through the 1 Ohm resistor, I(2) denote the current through the 2 ohm resistor, and so on.  The total current leaving point A is denoted by I.  We need to use Kirchhoff's
rules to find 7 equations for the seven currents, and then use algebra to solve those seven equations simultaneously.
(1) For the junction labeled J1 we have I(1) +I(2) +I(5)-I=0.
(2) For the junction labeled J2 we have I(2)-I(7)-I(9)=0
(3) For the junction labeled J3 we have I(13)-I(7)-I(5)=0
(4) For loop 1 we have V-1*I(1)=0
(5) For loop 2 we have 1*(I1)-2*I(2)-9*I(9)=0
(6) For loop 3 we have 9*I(9)-7*I(7)-13*I(13)=0
(7) For loop 4 we have 2*I(2)+7*I(7)-5*I(5)=0

Let us now use equation 4 to eliminate I(1) from the other equations.
I(1)=V.  We now have six equations.
(1) For the junction labeled J1 we have V +I(2) +I(5)-I=0.
(2) For the junction labeled J2 we have I(2)-I(7)-I(9)=0
(3) For the junction labeled J3 we have I(13)-I(7)-I(5)=0
(5) For loop 2 we have V-2*I(2)-9*I(9)=0
(6) For loop 3 we have 9*I(9)-7*I(7)-13*I(13)=0
(7) For loop 4 we have 2*I(2)+7*I(7)-5*I(5)=0

Let us now use equation 2 to eliminate I(2) from the other equations.
I(2)=I(7)+I(9).  We now have five equations.
(1) For the junction labeled J1 we have V +I(7)+I(9)+I(5)-I=0.
(3) For the junction labeled J3 we have I(13)-I(7)-I(5)=0
(5) For loop 2 we have V-2*I(7)-11*I(9)=0
(6) For loop 3 we have 9*I(9)-7*I(7)-13*I(13)=0
(7) For loop 4 we have 9*I(7)+2*I(9)-5*I(5)=0

Let us now use equation 3 to eliminate I(5) from the other equations.
I(5)=I(13)-I(7).  We now have 4 equations.
(1) For the junction labeled J1 we have V +I(9)+I(13)-I=0.
(5) For loop 2 we have V-2*I(7)-11*I(9)=0
(6) For loop 3 we have 9*I(9)-7*I(7)-13*I(13)=0
(7) For loop 4 we have 14*I(7)+2*I(9)-5*I(13)=0

Let us now use equation 5 to eliminate I(7) from the other equations.
2*I(7)=V-11*I(9).  We now have three equations.
(1) For the junction labeled J1 we have V +I(9)+I(13)-I=0.
(6) For loop 3 we have -7*V+95*I(9)-26*I(13)=0
(7) For loop 4 we have 7*V-75*I(9)-5*I(13)=0

Let us now use equation 7 to eliminate I(9) from the other equations.
75*I(9)=7*V-5*I(13).  We now have 2 equations.
(1) For the junction labeled J1 we have 82*V+70*I(13)-75*I=0.
(6) For loop 3 we have 140*V-2425*I(13)=0

Let us now use equation 6 to eliminate I(13) from the other equations.
485*I(13)=28*V.  We now have 1 equations.
(1) For the junction labeled J1 we have 41730*V-36375*I=0.

I=(41730/36375)*V=1.147*V
R=V/I=0.87 ohms

The resistance is 0.87 ohms.

Exercise  (You can earn up to 5 points extra credit by completing this exercise.)

Links to other Web material:

Current and resistance
Multi-loop circuits

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