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Over the years, electrical circuits have becomeincreasingly complex, with more and morecomponents combining to achieve very preciseresults ( Figure 14.1 ). Such circuits typically include

power supplies, sensing devices, potential dividersand output devices. At one time, circuit designerswould start with a simple circuit and graduallymodify it until the desired result was achieved. This

is impossible today when circuits include manyhundreds or thousands of components.Instead, electronic engineers ( Figure 14.2 )

rely on computer-based design software which

Circuit design

Figure 14.1 A complex electronic circuit thisis the circuit board which controls a computershard drive.

Figure 14.2 A computer engineer in Californiauses a computer-aided design (CAD) softwaretool to design a circuit which will form part of amicroprocessor, the device at the heart ofevery computer.

can work out the effect of any combinationof components. This is only possible becausecomputers can be programmed with the equationswhich describe how current and voltage behave ina circuit. These equations, which include Ohmslaw and Kirchhoffs two laws, were establishedin the 18th century, but they have come intotheir own in the 21st century through their use in

computer-aided design (CAD) systems.

Revisiting Kirchhoffs rst lawThis law has already been considered in Chapter 9 .It relates to currents at a point in a circuit, and stemsfrom the fact that electric charge is conserved.Kirchhoffs rst law states that:

The sum of the currents entering any point (or junction) in a circuit is equal to the sum of thecurrents leaving that same point.

As an equation, we can write Kirchhoffs rst law as:

I in = I out

Here, the symbol (Greek letter sigma ) means thesum of all, so I in means the sum of all currentsentering into a point and I out means the sumof all currents leaving that point. This is the sortof equation which a computer program can use to

predict the behaviour of a complex circuit.

Kirchhoffs laws

Chapter 14

e-Learning

Objectives
http://coas_p1_ch09_it.pdf/http://coas_p1_ch09_it.pdf/

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Chapter 14: Kirchhoffs laws

152

10

I I

30

6.0 V+ +

2.0 V

loop

Figure 14.6 A circuit with two opposing batteries.

Kirchhoffs second lawThis law deals with e.m.f.s and voltages in a circuit.We will start by considering a simple circuit whichcontains a cell and two resistors of resistances R1 and

R2 (Figure 14.5 ). Since this is a simple series circuit,the current I must be the same all the way around,and we need not concern ourselves further withKirchhoffs rst law. For this circuit, we can writethe following equation:

E = IR1 + IR2

e.m.f. of battery = sum of p.d.s across the resistors

SAQ

1 Calculate I in and I out in Figure 14.3 . IsKirchhoffs rst law satis ed?

You should not nd these equations surprising.However, you may not realise that they are aconsequence of applying Kirchhoffs second law tothe circuit. This law states that:

3.0 A

4.0 A

2.5 A

0.5 A

1.0 A2.0 A

Figure 14.3 For SAQ 1 .

2 Use Kirchhoffs rst law to deduce the valueand direction of the current I x inFigure 14.4 .

3.0 A

7.0 A

P

2.0 A

I x


R 1 R 2

I I

E

loop

Figure 14.5 A simple series circuit.

The sum of the e.m.f.s around any loop in a circuit

is equal to the sum of the p.d.s around the loop.

You will see later ( page 155 ) that Kirchhoffs secondlaw is an expression of the conservation of energy.

We shall look at another example of how this lawcan be applied, and then look at how it can be appliedin general.

Figure 14.6 shows a circuit with two batteries(connected back-to-front) and two resistors. Again,the current is the same all the way round the circuit.

Using Kirchhoffs second law, we can nd thevalue of the current I . First, we calculate the sumof the e.m.f.s, taking account of the way that the

batteries are connected together:

sum of e.m.f.s = 6.0 V 2.0 V = 4.0 V

Answer

Answer

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Second, we calculate the sum of the p.d.s:

sum of p.d.s = ( I 10) + ( I 30) = 40 I

Equating these gives:

4.0 = 40 I

and so I = 0.1 A. No doubt, you could have solvedthis problem without formally applying Kirchhoffssecond law.

SAQ

3 Use Kirchhoffs second law todeduce the p.d. across the resistorof resistance R in the circuit shown in Figure 14.7 ,and hence nd the value of R. (Assume the

battery of e.m.f. 10 V has

negligible internal resistance.)

20

10 V

0.1 A

R

Figure 14.7 Circuit for SAQ 3 .

Applying Kirchhoffs lawsFigure 14.8 shows a more complex circuit, with morethan one loop. Again there are two batteries andtwo resistors. The problem is to nd the current inthe resistors. There are several steps in this; Workedexample 1 shows how such a problem is solved.

6.0 V

30

2.0 V

I 3

I 2 I 2

I 1

P

I 1

10

Figure 14.8 Kirchhoffs laws are needed todetermine the currents in this circuit.

Calculate the current in each of the resistors in thecircuit shown in Figure 14.8 .

Step 1 Mark the currents owing. The diagramshows I 1, I 2 and I 3; note that it does not matter ifwe mark these owing in the wrong directions, asthey will simply appear as negative quantities inthe solutions.

Step 2 Apply Kirchhoffs rst law. At point P,this gives:

I 1 + I 2 = I 3 (1)

Step 3 Choose a loop and apply Kirchhoffssecond law. Around the upper loop, this gives:

6.0 = ( I 3 30) + ( I 1 10) (2)

Step 4 Repeat step 3 around other loops until thereare the same number of equations as unknowncurrents. Around the lower loop, this gives:

2.0 = I 3 30 (3)

We now have three equations with threeunknowns (the three currents).

Worked example 1

continued

Hint

Answer

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154

SAQ

4 You can use Kirchhoffs second law to nd thecurrent I in the circuit shown in Figure 14.10 .Choosing the best loop can simplify the problem.a Which loop in the circuit should you

choose?b Calculate the current I .

Signs and directionsCaution is necessary when applying Kirchhoffs secondlaw. You need to take account of the ways in which thesources of e.m.f. are connected and the directions of the

currents. Figure 14.9 shows a loop from a complicatedcircuit to illustrate this point. Only the components andcurrents within the loop are shown.

Step 5 Solve these equations as simultaneousequations. In this case, the situation has beenchosen to give simple solutions. Equation 3gives I 3 = 0.067 A, and substituting this value inequation 2 gives I 1 = 0.400 A. We can now nd I 2

by substituting in equation 1:

I 2 = I 3 I 1 = 0.067 0.400 = 0.333 A 0.33 A

Thus I 2 is negative it is in the opposite directionto the arrow shown in Figure 14.7 .

Note that there is a third loop in this circuit; wecould have applied Kirchhoffs second law to theoutermost loop of the circuit. This gives a fourthequation:

6 2 = I 1 10

However, this is not an independent equation; wecould have arrived at it by subtracting equation 3from equation 2.

e.m.f.sStarting with the cell of e.m.f. E 1 and workinganticlockwise around the loop (because E 1 ispushing current anticlockwise):

sum of e.m.f.s = E 1 + E 2 E 3

Note that E 3 is opposing the other two e.m.f.s.

p.d.sStarting from the same point, and workinganticlockwise again:

sum of p.d.s = I 1 R1 I 2 R2 I 2 R3 + I 1 R4

Note that the direction of current I 2 is clockwise, sothe p.d.s that involve I 2 are negative.

I 1

R 4

I 2

R 3 R 2

I 2

I 1

R 1

E 2

E 1

E 3

Figure 14.9 A loop extracted from a complicatedcircuit.

5.0 V

5.0 V

5.0 V

20

10

I

2.0 V

10

Figure 14.10 Careful choice of a suitable loopcan make it easier to solve problems like this.

Extension

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5 Use Kirchhoffs second law todeduce the resistance R of theresistor shown in the circuit loopof Figure 14.11 .

10

R

10 V

10

20

30 V

0.5 A

0.2 A

Conservation of energyKirchhoffs second law is a consequence of the

principle of conservation of energy. If a charge, say1 C, moves around the circuit, it gains energy as itmoves through each source of e.m.f. and loses energyas it passes through each p.d. If the charge moves all

the way round the circuit, so that it ends up whereit started, it must have the same energy at the endas at the beginning. (Otherwise we would be able tocreate energy from nothing simply by moving chargesaround circuits.) So:

energy gained passing through sources of e.m.f. = energy lost passing through components with p.d.s

You should recall that an e.m.f. in volts is simply theenergy gained per 1 C of charge as it passes through asource. Similarly, a p.d. is the energy lost per 1 C as it

passes through a component.

1 volt = 1 joule per coulomb


Hence we can think of Kirchhoffs second law as:

energy gained per coulomb around loop = energy lost per coulomb around loop

Here is another way to think of the meaning of e.m.f.A 1.5 V cell gives 1.5 J of energy to each coulomb ofcharge which passes through it. The charge then movesround the circuit, transferring the energy to componentsin the circuit. The consequence is that, by driving 1 Cof charge around the circuit, the cell transfers 1.5 J ofenergy. Hence the e.m.f. of a source simply tells us theamount of energy (in J) transferred by the source indriving unit charge (1 C) around a circuit.

SAQ

6 Use the idea of the energy gained and lost by a 1 C

charge to explain why two 6 V batteries connectedtogether in series can give an e.m.f. of 12 V or 0 V, but connected in parallel theygive an e.m.f. of 6 V.

7 Apply Kirchhoffs laws to thecircuit shown in Figure 14.12 to determine the current that will

be shown by the ammeters A 1,A2 and A 3.

20

10 V

20 5.0 V

A 3

A 1

A 2

Figure 14.12 Kirchhoffs laws make it possible todeduce the ammeter readings.

Hint

Answer

Answer

Hint

Answer

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156

If we apply Kirchhoffs second law to the loop thatcontains the two resistors, we have:

I 1 R1 I 2 R2 = 0 V

(because there is no source of e.m.f. in the loop). Thisequation states that the two resistors have the same

p.d. V across them. Hence we can write:

I =V

R

I 1 =V

R1

I 2 =V

R2

Substituting in I = I 1 + I 2 and cancelling the commonfactor V gives:

1 R

=1

R1 +

1 R2

For three or more resistors, the equation for totalresistance R becomes:

1

R =

1

R1 +

1

R2 +

1

R3 +

R 1 R 2 I I

V 1 V 2

V

Figure 14.13 Resistors in series.

Resistors in parallelFor two resistors of resistances R1 and R2 connectedin parallel ( Figure 14.14 ), we have a situation wherethe current divides between them. Hence, usingKirchhoffs rst law, we can write:

I = I 1 + I 2

R 1

R 2

I 2

I 1

I V

I

Figure 14.14 Resistors connected in parallel.

Resistor combinationsYou are already familiar with the formulae used tocalculate the combined resistance R of two or moreresistors connected in series or in parallel. Toderive these formulae we have to make use ofKirchhoffs laws.

Resistors in seriesTake two resistors of resistances R1 and R2 connectedin series ( Figure 14.13 ). According to Kirchhoffs

rst law, the current in each resistor is the same. The p.d. V across the combination is equal to the sum ofthe p.d.s across the two resistors:

V = V 1 + V 2

Since V = IR, V 1 = IR1 and V 2 = IR2, we can write:

IR = IR1 + IR2

Cancelling the common factor of current I gives:

R = R1 + R2

For three or more resistors, the equation for totalresistance R becomes:

R = R1 + R2 + R3 +

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9 Apply Kirchhoffs laws to nd the current at pointX in the circuit shown in Figure 14.16 .What is the direction of thecurrent?


4.0 V

X

10 V

20

80

20

Figure 14.16 For SAQ 9 .A

5.0 V

30

I

60

SAQ

8 There are two ways to calculate the current I inthe ammeter in Figure 14.15 . Both should give thesame answer.a Apply Kirchhoffs laws to determine the

current I .b Calculate the total resistance R of the two

parallel resistors, and hencedetermine the current I .

Summary

Kirchhoffs rst law represents the conservation of charge at a point in a circuit: sum of currents entering a point = sum of currents leaving that point Kirchhoffs second law represents the conservation of energy in an electric circuit:

sum of all the e.m.f.s around a circuit loop = sum of all the p.d.s around that loop

Answer

Answer

Glossary

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158

Questions1 a The statement of Kirchhoffs second law is based on which conservation law? [1]

120

15 V

0.08 A

X

b In the circuit above, determine: i the p.d. across the resistor X in the circuit [3] ii the resistance R of the resistor labelled X. [2]

[Total 6]

2 a State Kirchhoffs rst law. [2]

b Apply Kirchhoffs laws to the circuit below to determine the current I at point A inmilliamperes (mA). [4]

30 10

3.0 V

A

9.0 V [Total 6]

Hint

Answer

Hint

Hint

Answer

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COAS_P1_Ch14_it