Topic 5 Circuit Analysis

8/13/2019 Topic 5 Circuit Analysis

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INTRODUCTION

In this topic we will investigate d.c. circuits comprising resistors connected in

series and in parallel. These circuits can often be simplified by using the series-

parallel reduction method. However, there are also circuits which cannot besimplified using this method. To deal with such circuits we have to apply

Kirchoffs laws. In our discussions, we will assume that the batteries have no

internal resistance.

RESISTORS IN SERIES

Resistors connected in series have the same current flowing in each of them. Anexample of a series circuit is shown in Figure 5.1

5.1

TTooppiicc55 CircuitAnalysisLEARNING OUTCOMES

By the end of this topic, you should be able to:

1. Calculate the equivalent resistance of a group of resistors using theseries-parallel reduction method; and

2. Apply Kirchoffs laws to solve multiloop circuits.


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TOPIC 5 CIRCUIT ANALYSIS 69

Figure. 5.1: Resistors in series

The potential difference across each resistor is given by Ohms law. That is,

33

22

11

IRV

IRVIRV

(5.1)

The potential difference across the resistors is commonly referred to as the voltage

drop. Figure 5.1 shows the polarity of voltage drops. The current always flowsinto the positive terminal of a resistance. The potential difference across a battery

is called the electromotive force, emf, and its polarity is such that the current

flows out of the positive terminal.

The sum of voltage drops in a series circuit is equal to the battery emf:

eqIR

RRRI

IRIRIR

VVVE

)( 321

321

321

(5.2)

where the equivalent resistance, eqR , is defined as:

1 2 3eqR ( R R R ) (5.3)

The equivalent resistance of resistors connected in series is equal to the sum oftheir individual resistances.


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TOPIC 5 CIRCUIT ANALYSIS70

Thus the current in a series circuit is given by:

eqR

EI (5.4)

RESISTORS IN PARALLEL

Resistors connected in parallel have the same voltage drop across each resistor.

An example of a parallel circuit is shown in Figure. 5.2.

Figure 5.2: Resistors in parallel

5.2

What is the current and potential difference across each resistor in

Figure. 5.1, if 50,150,200 321 RRR and VE 10 ?

An interesting interactive exercise can be found at the URL shown below.It is a Java applet showing a circuit with two resistors in series:

http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/resist2/

index.html

The current in resistors connected in series is the same. The total voltage is thesum of the voltages across each resistor.

ACTIVITY 5.1

EXERCISE 5.1


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The voltage drop across each resistor is equal to the battery emf, E. The current in

each battery is:

3

3

2

2

1

1

R

EI

R

EI

R

EI

(5.5)

The sum of currents in a parallel circuit is equal the current provided by thebattery:

eqR

E

RRRE

R

E

R

E

R

E

IIII

321

321

321

111 (5.6)

For parallel resistors, the reciprocal of equivalent resistances eqR is equivalent to

the sum of the reciprocals of individual resistances.

321

1111

RRRReq (5.7)

What is the potential difference across each resistor and total current in

Figure 5.2, if 50,150,200 321 RRR and VE 10 .

The voltage across each resistor connected in parallel is the same. The total

current is the sum of the currents through each resistor.

EXERCISE 5.2


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Example 5.1:

What is the equivalent resistance between points A and B of the following circuit?

Solution:

The two 10 resistors are in series. Their equivalent resistance is 20 . Thecircuit can be simplified to Figure 5.3(a).

Figure 5.3(a) Figure 5.3(b)

The two 20 resistors in Figure 5.3(b) are in parallel. Their equivalent resistanceis

20

1

20

11

eqR

10eqR

The circuit of Figure 5.3(a) is simplified to Figure 5.3(c). The two 10 resistorsare in series. Their equivalent resistance is 20 which is in parallel with another20resistor as in Figure 5.3(c).

20

10

10 10

20

A

B


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Figure 5.3(c) Figure 5.3(d)

Finally the equivalent resistance of two 20 resistors in parallel is 10. Theequivalent resistance of the circuit is 10shown in Figure 5.3(d).

CIRCUIT ANALYSIS USING THE SERIES- PARALLEL REDUCTION METHOD

Often an electric circuit consists of combinations of both series and parallel

connections as shown below. In such cases, we can apply the reduction method toreduce the original circuit to one equivalent resistance. Lets see how this is done

in Example 5.2.

Example 5.2:

For the circuit shown Figure 5.4(a) below, find the equivalent resistance using theseries-parallel reduction method.

5.3

Find the equivalent resistance of the circuit below.

4

4

2

2

10

10

EXERCISE 5.3


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Figure 5.4(a)

Solution:

The current and voltage in each resistor is obtained by applying Equation (5.3)

and Equation (5.7).

The 250and 150are in series so that their equivalent resistance is 400. Thecircuit can be simplified to the equivalent circuit shown in Figure. 2.4b.

Figure 5.4b

In Figure 5.4(b), the 400resistor is in parallel with the 180. Their equivalentresistance is 124.

The circuit in Figure 5.4(b) can be simplified to a circuit containing three resistorsin series as in Figure 5.4(c). Finally, the circuit is reduced to its equivalent circuit

as in Figure 5.4(d).


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Figure 5.4(c) Figure 5.4(d)

Example 5.3

For the previous example, find the current in each resistor.

Solution:

Refer to Figure 5.4(d). The total current is:

24= = 0.054A

444I

The current in the 100and 220resistors are 0.054A.

The voltage between point A and B is VAB= 124x 0.054A = 6.7 V.

The current through 180resistor is AVAB 037.0

180

7.6

180

The current through 150and 250resistors is AVAB 017.0

400

7.6

400

To test your understanding, answer the following exercise.


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KIRCHOFF'S RULES

The circuit analysis above combined resistors in series and parallel. However,

there are many circuits in which the resistors are neither in series nor in paralleland as a result we are unable to use the series-parallel reduction method. To deal

with such circuits it is necessary to employ Kirchoff's rules.

There are two rules, namely, Kirchoff's junction rule and Kirchoff's loop rule.

5.4.1 Kirchoff's junction ruleKirchoffs junction rule states that the algebraic sum of currents at any junction iszero. Figure 5.5 shows three wires connected to form a junction or node.

Figure 5.5

5.4

Find the total current supplied by the battery in the circuit.

2

4

2

4 4

12V

2

EXERCISE 5.4


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Applying Kirchoff's junction rule, the algebraic sum of the current is

0321 III (5.7)

In applying the rule to a junction, the current that enters a junction is assigned apositive sign and the current that leaves the junction is assigned a negative sign.

As an example let us consider the circuit below which consists of two junctions a

and b;

Figure 5.5

By applying Kirchoffs junction rule at junction a, we obtain:

321

321 0

IIII

IIII

(5.8a)

Similarly, applying Kirchoffs junction rule at junction b gives:

321

321 0

IIII

IIII

(5.8b)

From Equation (5.8b) it is clear that the sum of currents in a parallel circuit is

equal to the current provided by the battery in Equation (5.6).

Kirchoffs junction rule states that the sum of currents entering a junction is

equal the sum of currents leaving the junction.


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A detailed explanation of Kirchoffs junction rule is given here:http://www.facstaff.bucknell.edu/mastascu/eLessonsHTML/Basic/Basic4Ki.html

A simple simulation that demonstrates the principle of conservation of charge isincluded, as well as problems to test your understanding.

5.4.2 Kirchoff's loop rule

Kirchoff's loop rule states that the algebraic sum of potential differences around

any complete loop of a circuit is zero.

Figure 5.7

Applying Kirchoff's loop rule in the loop of Figure 5.7, we obtain:

VVVE

VVVE

21

321 0 (5.9)

The sum of voltage drops in a series circuit is equal to the battery emf, similar to

Equation (5.2).

Kirchoff's loop rule states that the sum of potential difference around closedloop is equal zero.


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A detailed explanation of Kirchoffs loop rule is given here:http://www.facstaff.bucknell.edu/mastascu/eLessonsHTML/Basic/Basic5Kv.html

The explanation is simple, with interactive simulations and concept questions.

Note:

The general rule is that a rise or an increase in electric potential should be taken aspositive whereas a fall or a decrease should be taken as negative. In problems

where we have to apply Kirchoffs rules , the following situations arise:

(i) Potential drop across a resistorR

Lets traverse the resistor having resistanceRfrom point ato point b.

The direction of the current, I will indicate which point is at a higherpotential.

In Figure 5.9 (a), the resistor is traversed in the same direction as the

current. Thus the change in the electric potential (or the voltage drop)across the resistor is IR.

However, in Figure 5.9 (b), the resistor is traversed in the direction

opposite to the current. The change in the electric potential across theresistor is now +IR.

Figure 5.9: The potential drop across a resistor

(ii) Potential drop across a battery of emf

If we is traverse in the direction of the battery (ie from to +), then

change in the electric potential is +. See Figure 5.9(a)

However in Figure 5.9(d), if we traverse in the direction opposite to the

battery (ie from + to ), the change in the electric potential is now


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Figure 5.9: The potential drop across a battery

The technique of applying Kirchoffs junction and Kirchoffs loop rules to a

circuit can be illustrated with a few examples.

Example 5.4:

Determine the currentI in the circuit.

Figure 5.10


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Solution:

The first step is to draw the current in an arbitrary direction. The second step is tomark the polarity of the potential difference across each resistor.

Applying Kirchoffs loop rule to the circuit, from Equation. (5.9) we obtain

AI

III

13.0

030125010036

The current is a positive number, indicating the initial choice for direction ofcurrent was correct.

Example 5.5:

Consider the circuit shown in Figure 5.11 below

(a) Apply the junction rule to point b

(b) Use Kirchoffs Loop Rule to obtain the equation for loop abefa

(c) Use Kirchoffs Loop Rule to obtain the equation for loop bcdeb

(d) Obtain the values of 1 2,I I and 3I .

Figure 5.11

Solution:The first step is to draw the loop currents in arbitrary directions. The second step

is to mark the polarity of potential difference across each resistor. Since there are

three unknown currents, we need to apply Kirchoffs current rule and loop rule to

obtain three simultaneous equations.


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(a) The junction rule when applied to junction b gives

1 2 3 0................................... .1I I I Eq

(b) Applying Kirchoffs Loop Rule to loop abefa gives

3 1

1 3

1 3

12 200 100 6 0

100 200 6

2 0.06....................................... 2

V I I

I I

I I Eq

(c) Applying Kirchoffs Loop Rule to loop bcdeb gives

2 3

2 3

1 3 3

1 3 3

1 3 3

1 3

1 3

50 200 12 0

50 200 12

50 200 12

50 50 200 12

50 50 200 12

50 250 12

5 0.24................................... .3

I I V

I I

I I I

I I I

I I I

I I

I I Eq

(d) The last two equations can be solved simultaneously to obtain2

I and 3I .

Eq.3 + Eq.2

3

3

7 0.30

0.043

I

I A

From Eq. 2,

1 32 0.06

2 0.043 0.06

0.026

I I

A

Finally, from Eq. 1

2 3 1I I I = 0.043-0.026= -0.069 A


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The negative answer for 2I indicates that the current through 50 resistor is not

directed from right to left as drawn but instead the current actually flows from left

to right.

We have analysed d.c circuits with more than one loop using two methods,namely the series-parallel reduction method and Kirchoffs rules.

The current in resistors connected in series is the same. The total voltage isthe sum of the voltages across each resistor.

The voltage across each resistor connected in parallel is the same. The totalcurrent is the sum of the currents through each resistor.

Kirchoffs junction rule states that the sum of currents entering a junction is

equal the sum of currents leaving the junction.

Kirchoffs loop rule states that the algebraic sum of potential differencesaround any complete loop of a circuit is zero.

5

10

4

10V

5V

a

b

3I

2I

1I

1. Determine the currents 1I , 2I and 3I :

EXERCISE 5.5


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Equivalent resistance

Kirchoffs junction rule

Kirchoffs loop rule

1. In the circuit shown below, the current1

I is 1.5 A.

Find 2I , 3I and 2V .

2. Three resistors, of 10 each, are connected in parallel. The combination isthen connected in series with 2 V battery of 1 resistance. Find the totalbattery current.

1. Find the equivalent resistance between a and b below:

100

20

40

b

a


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2. A 10 resistor is connected in parallel with a 20 resistor. A p.d is nowapplied to this combination.

(a) Which resistor will carry the greater current?

(b) What will the ratio of the currents be?

Topic 5 Circuit Analysis

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