Lecture 8

* Electromotive Force.

*Resistors in Series and in Parallel.

* Kirchhoff 's Rules.

* Electrical Instruments.

Electromotive Force

*A constant current can be maintained in a closed circuits through the use of a source of energy, called an electromotive force (emf).

* A source of emf is any device ( such as battery or generator) that will increase the potential energy of charges circulating in a circuit.

* The emf, ε , of a source describes the work done per unit charge.

* The SI units of emf is the volt.

*consider the circuit as shown in the figure-72,

the battery within the dotted rectangle is

represented by a source of emf, ε, in series with

the internal resistance, r, as the charge passes

from negative to the positive terminal of the

battery, its potential increase by ε. However, as

it moves through the resistance, r, its potential

decreases by a mount, Ir, where I the current in

the circuit.

* The terminal voltage of

the battery is given by:

•*From the figure-72, we see that the terminal

voltage (V) must equal the potential difference

across the external resistance, R, often called the

load resistance.

•* That is, V=IR, combining this with the last

equation, we see that:

•*The current gives:

•* If we multiply the equation by the current I, we find that the total power output of the source of emf, Iε, is converted into power dissipated as heat in the load resistance, plus the power dissipated in the internal resistance, :

Example-1:

•* A battery has an emf of 12 volt and the internal

resistance of 0.05 Ω, its terminals are connected

to a load resistance of 3 Ω:

•a- find the current in the circuit and the terminal

voltage of the battery?

•b- find the power dissipated by the internal

resistance? The power dissipated in the load

resistance? The power delivered by the battery?

Resistors in series and in parallel 1-Series Combination: * For a series connection of resistors, the current

is the same in each resistor. Since any charge flows

through must equal the charge flows through .

* From the figure-73, the potential drop from (a)

to (b) equal and the potential drop from (b) to

(c) equal .

* The potential drop from (a)

to (c) is given by:

* We can replace the two resistors in series by a

single equivalent resistance whose the value

is the sum of the individual resistance:

* The equivalent resistance of three or more

resistors connected in series:

* The equivalent resistance of a series connection of resistors is always greater than any individual resistance.

2-Parallel Combination:

* Consider two resistors connected in parallel as

shown in figure-74.

* There is an equal potential across each

resistor and the current in each resistor not the

same.

*When the current (I) reaches point (a), it splits

in two parts, going through and going

through .

* If is greater than , then will be less than

, that is the charge will tend to take the path of

least resistance.

* Clearly, since charge must be conversed, the

current(I) that enters point (a) must equal the

total current leaving this point:

* From Ohms law:

•*The equivalent resistance of two resistors in

parallel:

•*The equivalent resistance of three or more

resistors in parallel:

* The equivalent resistance of a parallel

connection of resistors is always less than the

smallest resistance in the group.

Kirchhoff's Law:.

* Simple circuits can be analyzed using ohms

law and the rules for series and parallel

combinations of resistors.

*Often it is not possible to reduce a circuit to a

single loop. The procedure for analyzing more

complex circuits is greatly simplified by the use

of two simple rules called Kirchhoff's Rules.

Kirchhoff's Rules:.

1- The sum of the currents entering any junction

must equal the sum of the currents leaving that

junction.(A junction is any point in the circuit

where a current can split.

2- The algebraic sum of the changes in potential

across all of the elements around any closed

circuit loop must be zero.

* The first law is a statement of

conservation of charge, if we

apply this rule on the junction as

shown in figure-75, we get:

* The second rule follows from conservation

of energy.

* That is, any charge moves around any closed

loop in a circuit (it starts and ends at the

same point) must gain as much energy as it

loses.

*As an aid in applying the second rule, the

following calculation tools should be noted:

1- If a resistor is traversed in the direction of

the current, the change in potential across the

resistor is(-IR). (fig-76a)

2- If a resistor is traversed in the direction

opposite the current, the change in potential

across the resistor is(+IR). (fig-76b)

3- If a source of emf is traversed in the

direction of the emf (from – to + on the

terminals), the change in potential is(+ε). (fig-

76c)

4- If a source of emf is

traversed in the direction

opposite the emf

(from + to - on the terminals),

the change in potential is(-ε).

(fig-76d)

1-Example

•A single loop circuit contains two external resistors

and two sources of emf as shown in figure-77, the

internal resistances of the batteries have been

neglected.

• a- Find the current in the circuit?

• b- The power lost in each resistor?

Example-2

•Find the currents , and in the circuits shown

in figure-78.

Home work( a multiloop circuit) •The multiloop circuit in figure-79, contains three

resistors, three batteries and one capacitor, Under

steady – state conditions, find the unknown current?

Analyze this circuit for me, please. Find the currents I1, I2, and I3.

1 1 = 85 V

1 2 = 45 V

20

30

40

a b c

d

e g f

h

I3

I2

I1

Kirchhoff’s Rules

1 1 = 85 V

1 2 = 45 V

20

30

40

a b c

d

e g f

h

I3

I2

I1

I see two sets of resistors in series. This. And this.

You know how to analyze those.

Further analysis is difficult. For example, series1 seems to be in parallel with the 30

resistor, but what about 2? We haven’t discussed how to analyze that combination.

series1

series2

A new technique is needed to analyze this, and far more complex circuits.

Kirchhoff ’s Rules

Kirchhoff ’s Junction Rule: at any junction point, the sum of all currents entering the junction

must equal the sum of all currents leaving the junction. Also called Kirchhoff ’s First Rule.*

Kirchhoff ’s Loop Rule: the sum of the changes of potential around any closed path of a

circuit must be zero. Also called Kirchhoff ’s Second Rule.**

*This is just conservation of charge: charge in = charge out.

**This is just conservation of energy: a charge ending up where it started out

neither gains nor loses energy (Ei = Ef ).

Kirchhoff ’s Rules

Starting Equations

at any junctionI=0

around any closed loopV=0

simple… but there are details to worry about…

Solving Problems with Kirchhoff ’s Rules

Lit any for Circuit Problems

1. Draw the circuit.

2. Label + and – for each battery (the short side is -).

3. Label the current in each branch of the circuit with a symbol and an arrow. You

may choose whichever direction you wish for the arrow.

4. Apply Kirchoff ’s Junction Rule at each junction. The direction of the current arrows

tell you whether current is flowing in (+) or out (-).

Step 4 will probably give you fewer equations than variables. Proceed to step 5 go get

additional equations.

5. Apply Kirchhoff ’s Loop Rule for as many loops as necessary to get enough

equations to solve for your unknowns. Follow each loop in one direction only—your

choice.

5a. For a resistor, the sign of the potential difference is negative if your chosen

loop direction is the same as the chosen current direction through that

resistor; positive if opposite.

5b. For a battery, the sign of the potential difference is positive if your chosen

loop direction moves from the negative terminal towards the positive;

negative if opposite.

6. Collect equations, solve, and check results.

1. Draw the circuit.

2. Label + and – for each battery.

3. Label the current in each branch of the circuit with a symbol and an arrow (OK to

guess direction).

4. Apply Kirchhoff ’s Junction Rule at each junction. Current in is +.

5. Apply Kirchhoff ’s Loop Rule for as many loops as necessary. Follow each loop in one

direction only.

5a. Resistor: +

- I

loop

V is - 5b. Battery:

loop

V is +

6. Solve.

1 1 = 85 V

1 2 = 45 V

20

40

a b c

d

e g f

h

I3

I2

I1

Back to our circuit: we have 3 unknowns (I1, I2, and I3), so we will need 3 equations. We

begin with the junctions.

Junction a: I3 – I1 – I2 = 0 --eq. 1

Junction d: -I3 + I1 + I2 = 0

Junction d gave no new information, so we still need two more equations.

30

d a

1 1 = 85 V

1 2 = 45 V

20

40

a b c

d

e g f

h

I3

I2

I1

There are three loops. Loop 1. Loop 2. Loop 3.

Any two loops will produce independent equations. Using the third loop will provide no

new information.

30

Reminders:

I

loop

V is - + -

loop

V is +

The “green” loop (a-h-d-c-b-a):

(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0

- 30 I1 + 45 - 41 I3 = 0 --eq. 2

The “blue” loop (a-b-c-d-e-f-g):

(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0

41 I3 -130 + 21 I2 = 0 --eq. 3

Three equations, three unknowns; the rest is “algebra.”

Make sure to use voltages in V and resistances in . Then currents will be in A.

5

Collect our three equations:

I3 – I1 – I2 = 0

- 30 I1 + 45 - 41 I3 = 0

41 I3 -130 + 21 I2 = 0

Rearrange to get variables in “right” order:

– I1 – I2 + I3 = 0

- 30 I1 - 41 I3 + 45 = 0

21 I2 + 41 I3 -130 = 0

Use the middle equation to eliminate I1:

I1 = (41 I3 – 45)/(-30)

There are many valid sets of steps to solving a system of equations. Any that works is acceptable.

Two equations left to solve:

– (41 I3 – 45)/(-30) – I2 + I3 = 0

21 I2 + 41 I3 -130 = 0

Might as well work out the numbers:

1.37 I3 – 1.5 – I2 + I3 = 0

21 I2 + 41 I3 -130 = 0

– I2 + 2.37 I3 – 1.5 = 0

21 I2 + 41 I3 -130 = 0

Multiply the top equation by 21:

– 21 I2 + 49.8 I3 – 31.5 = 0

21 I2 + 41 I3 -130 = 0

Add the two equations to eliminate I2:

– 21 I2 + 49.8 I3 – 31.5 = 0

+ ( 21 I2 + 41 I3 -130 = 0 )

90.8 I3 – 161.5 = 0

Solve for I3:

I3 = 161.5 / 90.8

I3 = 1.78

Go back to the “middle equation” two slides ago for I1:

I1 = (41 I3 – 45)/(-30)

I1 = - 1.37 I3 + 1.5

I1 = - (1.37) (1.78) + 1.5

I1 = - 0.94

Go back two slides to get an equation that gives I2:

– I2 + 2.37 I3 – 1.5 = 0

I2 = 2.37 I3 – 1.5

I2 = (2.37) (1.78) – 1.5

I2 = 2.72

Summarize answers so your lazy prof doesn’t have to go searching for them and get

irritated (don’t forget to show units in your answer):

I2 = 2.72 A

I3 = 1.78 A

I1 = - 0.94 A

Are these currents correct? How could you tell? We’d better check our results.

I3 – I1 – I2 = 0

- 30 I1 + 45 - 41 I3 = 0

41 I3 -130 + 21 I2 = 0

I2 = 2.72 A

I3 = 1.78 A

I1 = - 0.94 A

1.78 – (-0.94) – 2.72 = 0 ?

- 30 (-0.94) + 45 - 41 (1.78) = 0.22 ?

41 (1.78) -130 + 21 (2.72) = 0.10 ?

Are the last two indication of a mistake or just round off error? Recalculating while

retaining 2 more digits gives I1=0.933, I2=2.714, I3=1.7806, and the last two results are

0.01 or less roundoff was the culprit.

Electrical Instruments

•*The Ammeter: • A device that measure the current , the current

to be measured must pass directly through the

ammeter so that the ammeter is in series with

current it is to measure.

•* An ideal ammeter should have zero resistance

so as not to alter the current being measured.

•* The Voltmeter: •A device that measures potential differences, the

potential difference between any two points in the

circuit can be measured by attaching the terminals

of the voltmeter between these points without

breaking the circuit.

• An ideal voltmeter has infinite resistance so that

no current will pass through it.

•* The Galvanometer: •The galvanometer is the main component used in

the construction of ammeter and voltmeter.

•* The Wheatstone bridge: • Unknown resistances can be measured using a

circuit known as a Wheatstone bridge (figure-83),

•This circuit consists of the unknown resistance

and three known resistance ,

, , , galvanometer

and a source of emf.

* The principle of its operation is quite simple,

the resistor is varied until the galvanometer

reading is zero, so there is no current from a

to b, under this condition the bridge is said to

be balanced. Since the potential at point a

must equal the potential at point b, the

potential across must equal the potential

across .

* Dividing 1 by 2 , we find:

* Likewise, the potential across must

equal the potential across :