Mod- 5 Dccurrentnew

CURRENTS AND OHMS LAW 1. Electric Current  Example: Problem 1 o 2. The resistivity of materials  Example: Problem 2  Example: Problem 3 o 3. Resistance in combination  Example: Problem 4 Example: Problem 5 o



CURRENTS AND OHMS LAW 1. Electric Current

Figure 1. Electric field in a wire. Electric current is the flow of free charge charier through a material. These free charge carriers can be free electrons, holes or ions. When a wire is connected to the terminals of a battery, an electric field is generated inside the wire (see Figure 1). The free electrons in the wire will move in a direction opposite to that of the field lines. The electric charge will try to redistribute itself in such a way that the net electric field in the wire is equal to zero. However, the positive terminal of a battery acts as a sink for electrons and the negative terminal acts as a source of electrons, and a continuous flow of electrons will be created. This continuous flow of electrons is called an electric current. The symbol of current is I and its SI unit is the Ampere (A). The current is defined as

I=

dq dt

…..

…..

……

(1)

where dq is the amount of charge that flows past some given point on the wire during a time period dt. A current of 1 A is equal to 1 C/s. The current density j is defined as

J =

I A

…..

……

……

(2)

where I is the current flowing through the conductor, and A is the cross-sectional area of the conductor. Even though the electrons feel an electric field inside the conductor, they will not accelerate. The electrons will experience significant friction as a result of collisions with the positive ions in the conductor. On average, the electrons will move with a constant speed from the negative terminal of the battery to the positive terminal. Their average velocity, called the drift velocity vd, is proportional to the electric field E vd ∝ E For a given density of free electrons in the conductor, an increase of the drift velocity of each of the electrons will increase the number of electrons passing by a given point on the conductor per unit of time. This is illustrated in Figure 2. During a time interval dt the electrons will cover, on average, a distance equal to dx where

dx = vd dt

Figure 2. Motion of average electron in conductor. All electrons within a distance dx from the point P will therefore pass this point during the time interval dt. Suppose the density of electrons in the conductor is n electrons/m3. The number of electrons dN that will pass P during the time interval dt is then equal to dN = n A vd dt ….. ….. ….. (5) Since each electron carries a charge e, the total charge dQ that will pass point P in a time interval dt is equal to dq = e dN = e n A vd dt …. ….. (6) The current through the conductor is therefore equal to

I=

dq = dt

e n A vd

…….

…….

…….

(7)

Equation (7) shows that the current in the conductor is proportional to the cross-sectional area of the conductor and proportional to the drift velocity. The above equation can be re-written as

I = n e vd A j = n e v d ……

or, Current density

…….

……

(8)

Since the drift velocity is proportional to the electric field E the following relation holds for the current in the conductor: 2 Ohm's Law “The current through a conductor at a constant temperature is proportional to the voltage applied across its terminal” I ∝ ∆V Or,

I=

∆V R

……..

……… ……..

Here R is the resistance of the conductor.

The resistance of a conductor is directly proportional to its length and inversely proportional to its area of cross-section so that

R=ρ

L A

…….

……..

……… (11)

the constant ρ is property of the material of the conductor and called its specific resistance or resistivity. 3. The resistivity of materials The resistivity ρ has as units ohm-meter (Ω . m). The resistivity of most conductors is between 10-8 Ω . m and 10-7 Ω . m. The resistivity of a conductor depends not only on the type of the material but also on its temperature. The resistivity of an insulator varies between 1011 Ω . m and 1017 Ω . m. whereas the semiconductor has resistivity 10-4 to 103 Ω . m In conductors the resistivity decreases with decreasing temperature. In some materials, such as lead, zinc, tin and niobium, the resistivity vanishes as the temperature approaches absolute zero. At these low temperatures, these materials exhibit superconductivity. Example: Problem 1 An aluminum wire has a resistance of 0.10 Ω. If you draw this wire through a die, making it thinner and twice as long, what will be its new resistance ? The initial resistance Ri of the aluminum wire with length L and cross-sectional area A is equal to

Ri = ρ

L A

…….

…..

….

(13)

The initial volume of the wire is L . A. After passing the wire through the die, it s length has changed to L' and its cross-sectional area is equal A'. Its final volume is therefore equal to L' A'. Since the density of the aluminum does not change, the volume of the wire does not change, and therefore the initial and final dimensions of the wire are related: L` A` = L A or

A′ =

LA L′

…..

…… …..

…… …..

……

(14)

(15)

The problem states that the length of the wire is doubled (L' = 2 L). The final cross-sectional area A' is therefore related to the initial cross-sectional area A in the following manner:

A′ =

A 2

…..

…..

…..

(16)

L′ L 2L = ρ = 4 ρ = 4 Ri ….. A A′ A 2

…..

…..

The final resistance Rf of the wire is given by Ω

Rf = ρ

(17)

The resistance of the wire has increased by a factor of four and is now 0.40 Ω.

Example: Problem 2 The air conditioner in a home draws a current of 12 A. Suppose that the pair of wires connecting the air conditioner to the fuse box are No. 10 copper wires with a diameter of 0.259 cm and a length of 25 m each. a) What is the potential drop along each wire ? Suppose that the voltage delivered to the home is exactly 110 V at the fuse box. What is the voltage delivered to the air conditioner ? b) Some older homes are wired with No. 12 copper wire with a diameter of 0.205 cm. Repeat the calculation of part (a) for this wire.

Figure 3. Wiring diagram of air conditioner in Problem 2. a) The resistivity of copper is 1.7 x 10-8 wire is equal to

RCu =

ρ Cu

.

m (see Table 1). The resistance RCu of each copper

L L = ρ Cu π 2 A d 4

…..

…..

…..

(18)

where L is the length of the wire and d is its diameter. A current I is flowing through the wires and I = 12 A. The voltage drop [Delta]V across each wire is equal to ∆V = I RCu = 4 I ρCu

L π d2

….

….

…..

(19)

Figure 3 shows schematically a wiring diagram of the air conditioner circuit. The voltage across the air-conditioner unit is equal to 110 - 2 . [Delta]V, where [Delta]V is given by eq.(19). The length of each copper cable is 25 m, and its diameter is equal to 0.259 cm. The voltage drop across each wire is thus equal to ∆V = 4 x 12 x 1.7 x 10-8 x

25

π ( 0.00259 )

2

= 0.95 V

…..

…..

(20)

The voltage across the AC unit is therefore equal to 108.1 V. b) A No. 12 wire has a diameter equal to 0.205 cm. The voltage drop across this wire is equal to ∆V = 4 x 12 x 1.7 x 10-8 x

25

π ( 0.00205)

2

= 1.55 V

…..

…..

(21)

and the voltage across the AC unit is equal to 106.9 V. Example: Problem 3 A high voltage transmission line has an aluminum cable of diameter 3.0 cm, 200 km long. What is the resistance of this cable ? The resistivity of aluminum is 2.8 x 10-8 m. the length of the cable is 200 km or 2 x 105 m. The diameter of the cable is 3 cm and its cross-sectional area is equal to [pi] (d/2) 2 or 7.1 x 10-4 m2. Substituting these values into eq.(11) the resistance of the cable can be determined

2 ×10 5 R = 2.8 × 10 × = 7.9 Ω ….. 7.1×10 −4 −8

…..

…..

(22)

4. Resistance in combination A device that is specifically designed to have a high resistance is called a resistor. The symbol of a resistor in a circuit diagram is a zigzag line (see Figure 4).

Figure 4. Symbol of a resistor. Figure 5 shows two resistors with resistance R1 and R2 connected in series. Suppose the current flowing through the circuit is equal to I. The voltage drop [Delta]V1 across resistor R1 is equal to ∆V1 = I R1 ….. ….. ….. …. (24)

and the voltage drop [Delta]V2 across resistor R2 is equal to ∆V2 = I R2

…..

…..

…..

….

(24)

The potential difference [Delta]V across the series circuit is equal to ∆V = ∆V1 + ∆V2 = I (R1 + R2) …..

….

(25)

Equation (25) shows that two resistors connected in series act like one resistor with a resistance equal to the sum of the resistance of resistor 1 and the resistance of resistor 2 R = R 1 + R2

…..

…..

…..

(26)

Figure 5. Two resistors connected in series. Figure 6 shows two resistors connected in parallel. In this circuit, the current through each resistor will be different, but the voltage drop [Delta]V across each resistor will be the same. Using Ohm's law the current I1 flowing through resistor R1 can be calculated

I1 =

∆V R1

….

….

….

(27)

….

(28)

and the current I2 flowing through resistor R2 is equal to

I2 =

∆V R2

….

….

The total current flowing through the circuit is equal to the sum of the currents through each resistor

I = I1 + I 2 =

 1 ∆V ∆V 1  ∆V + =  + R1 R2 R R  1 2 

…..

…..

(29)

The resistor network shown in Figure 6 is therefore equivalent to a single resistor R where R can be obtained from the following relation

1 1 1 = + R R1 R2

…..

….

…..

(30)

Equation (30) shows that the resistance of a parallel combination of resistors is always less than the resistance of each of the individual resistors.

Figure 6. Two resistor connected in parallel. Example: Problem 5 What is the resistance of the combination of four resistors shown in Figure 7. Each of the resistors has a value of R.

Figure 7. Problem 5. To find the net resistance of the circuit shown in Figure 7, we start calculating the net resistance R34 of the parallel circuit of resistors R3 and R4:

1 1 1 = + ……. R34 R3 R4 R34 =

or

R3 R4 …….. R3 + R4

……..

……. (37)

………

……. (38)

The circuit shown in Figure 7 is therefore equivalent with the circuit shown in Figure 8. Resistors R2 and R34 form a series network and can be replaced by a single resistor with a resistance R234 where

R234 = R2 + R34 ……..

………

Figure 8. Problem 5.

……..

(39)

Figure 9. Problem 5.

The circuit shown in Figure 28.8 can now be replaced by an equivalent circuit shown in Figure 28.9. The resistance Rtot of this circuit can be obtained from the following relation

1 1 1 = + Rtot R R234

or

Rtot =

R R234 R + R234

…….

……..

……. (40)

…….

……..

……..(41)

In the special case considered, R1 = R2 = R3 = R4 = R. Thus

RR 1 = R ……. R+ R 2 1 3 R234 = R + R = R ……. 2 2 3  R R  2  3 Rtot =  = R …… 3 R+ R 5 2 R34 =

…….

……... ……..

(42)

……… ……..

(43)

…….

(44)

For R = 3 Ω the total resistance is equal to 1.8 Ω.

DC CIRCUITS

1. Electromotive Force

Figure 1. Electron in electronic circuit. To keep a current flowing in an electronics circuit we need a source of electric potential. Consider the circuit shown in Figure 1. The electrons in the conductor move from the end with the low (negative) potential towards the end with the high (positive) potential. The velocity of the electron is limited by the friction it experiences while traveling through the conductor, and on average will not change. However, the electron will lose potential energy, and its total mechanical energy is therefore reduced when it arrives at the end of the circuit (positive terminal of the source). In order to sustain the continuous flow of electrons around the circuit, the source must force the electrons from the terminal with the positive potential to the terminal with the negative potential. During this process, the source does work on the electrons, and increases their total mechanical energy. The strength of the source is measured in terms of the electromotive force (emf). The emf of a source is defined as the amount of electric energy delivered by the source per Coulomb of positive charge as charge passes through the source from the low-potential terminal to the high-potential terminal. A steady current will flow through the circuit in Figure 1 if the increase of potential energy of the electrons in the source is equal to the change in the potential energy of the electrons along their path through the external circuit. The unit of emf is the Volt (V) and usually the emf is simply called the voltage of the source. The

symbol for emf is [epsilon]. The most important sources of emf are: Batteries: , Electric Generators:,Fuel cells: Solar Cell:etc. A source with a time-independent emf is represented by the symbol shown in Figure 2. The long line in Figure 2 represents the positive terminal of the source, while the short line represents the negative terminal.

Figure 2. Schematic symbol of source of emf.

4. Energy in circuits To keep a current flowing in a circuit, work must be done on the circulating charges. If a charge dq passes through a battery with emf [E], the work done dW will be equal to dW = E dq The rate at which the emf source does work is given by dW dq =E = Ei dt dt The rate of work is called the power P, and the unit of power is the Watt (W, 1 W = 1 VA). The moving charges dissipate some of their energy when passing through resistors. Suppose the potential drop across a resistor is ∆V. For an electron moving through the resistor the loss of potential energy is equal to e . ∆V. The energy lost is converted into heat, and the rate at which energy is dissipated is equal to

or, the conversion of electric energy into thermal energy is called Joule heating.

Problem

1 The current density in a cylindrical wire of radius R = 2.0mm is uniform across a cross section of the wire and is given by J = 2.0 × 10 5 A/m2. What is the current through the outer portion of the wire between radial distances R/2 and R? R R/2

2.

3.

4.

5.

For the above figure, suppose that the current density through a cross section varies with radial distance r as J = ar2 , in which a = 3.0 × 1011 A / m 4 and r is in meters. What now is the current through the outer portion of the wire between radial distances R/2 and R? One end of an aluminum wire whose diameter is 2.5 mm is welded to one end of a cooper wire whose diameter is 1.8 mm. the composite wire carries a steady current I of 17 mA. (a) What is the current density in each wire? (b) What is the drift speed of the conduction electrons in the copper wire? [ given that n = 8.47 × 10 28 electrons / m 3 ]. A strip of silicon has a rectangular cross section with width w = 3.2 mm and height h = 250 µm , and through which there is a uniform current I of 5.2 mA. [Consider n = 1.5 × 10 23 m −3 . (a) What is the current density in the strip? (b) What is the drift speed? A wire of length L = 2.35 m and diameter d = 1.63 mm carries a current I of 1.24 A. The wire dissipates electrical energy at the rate P of 48.5 mW. Of what is the wire made?