chapter 4

Figure 4.1

PHY193 - Basic Physics For Engineers II

CHAPTER 4 - CURRENT AND RESISTANCE

4.1 Electric current

- Electric current is defined as the net amount of charge passing per unit time through an area perpendicular to the direction of flow. The magnitude of current is rate of the net flow of charge. Figure 4.1 shows electric current flowing in a wire.

- If q is the net charge that passes through the shaded surface in Figure 4.1 during a time interval t, the current in the wire is defined as

I= ΔqΔt (4.1)

The SI unit for current is Ampere (A). Other unit is C s-1.

- According to convention, the direction of electric current is defined as the direction in which the positive charges are transported, that is opposite to the direction of flow of the electrons.

Example 4.1

Two wires of cross-sectional area 1.6 mm2 connect the terminals of a battery to the circuitry of the clock. During an interval of 0.04 s, 5.0 x 1014 electrons move to the right through a cross-section of one of the wires. What is the magnitude and direction of the current in the wire?

Solution

The magnitude of charge of 5.0 x 1014 electrons is

q = Ne = 5.0 x 1014 x 1.602 x 10-19 = 8.0 x 10-5 C

The magnitude of current is

I= ΔqΔt

=8 . 0×10−5

0 .04=0 . 002 A

Since electron move to the right, the direction of electric current is to the left.

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Figure 4.2

RL

r


4.2 Battery and Electromotive force (emf)

- One common type of supply of electron is the battery. A battery converts stored chemical potential energy into electrical energy. When a complete circuit is formed, electrons from the negative electrode (called cathode) will move through the wire to the positive electrode (called anode).

- The potential difference across the terminals of a battery when it is not connected to a circuit is the battery’s electromotive force or emf. The battery’s emf is defined as the work done by the battery per unit coulomb of charge that passes through it.

- When the battery is connected to a circuit and charges flow, the voltage across the terminal is slightly less that the emf. This voltage is known as the terminal voltage. The equation for terminal voltage is

V t=iR L (4.2)

where i = The current flowing in the circuit.RL = The load resistance of the circuit.

- Under many condition, the terminal voltage is almost the same as the emf. Slight difference between these two voltages is due to mainly the existence of internal resistance in the battery. The relation between these two voltages is given by

ξ−V t=ir (4.3)

where ξ = The emfVt = The terminal voltage r = The internal resistance of the battery

- Figure 3.2 shows the circuit diagram of a battery.

- It can be shown that

V t=( R Lr+RL )ξ (4.4)

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Example 4.2

A transistor radio battery has an emf of 9.0 V. When a short copper wire is connected directly across the battery terminals, a current of 4.0 A passes through the wire. a) What is the internal resistance of the battery?b) Determine i) the potential difference across a 10 load ii) the current flowing through the 10 load.

Solution

a) The resistance of the wire is negligibly small. The current is limited by the internal resistance of the battery. The internal resistance of the battery is

r=ξi=9 . 0

4 . 0=2 . 25Ω

b) When the battery is placed across a 10 load,

i) the potential difference, V t=(102 .25+10 )( 9. 0 )=7 .3V

ii ) the current, i =V tR L

=7 . 310

=0. 73 A

4.3 Resistance and resistivity

- The electrical resistance R is defined to be the ratio of the potential difference V across a conductor to the current I through the material.

R=ΔVI (4.5)

In SI units, electrical resistance is measured in ohms (). For a given potential difference, a large current will flow in a conductor with small resistance, while a small current flows through a conductor with a large resistance.

- Resistance depends on shape and size. The electrical resistance of a conductor of length L and cross-sectional area A is given by

R=ρ LA (4.6)

where R is proportional to the length of the conductor but inversely proportional to the cross-sectional area of the conductor. is the proportionality constant called the resistivity of the material. is an intrinsic characteristic of a particular material at a particular temperature.

Equation (4.6) assumes a uniform distribution of current across the cross-section of the conductor.

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(a)

R1 R3R2

i i

V(b)

i1

i4

i3

i2

R3

R2

R1

V

Figure 4.3


Example 4.3

What is the electrical resistance of an iron wire 0.50 m long with a diameter of 1.3 mm if the resistivity of the iron, iron

= 9.7 x 10-8 m?

Solution

The cross-sectional area of the wire,

A=πr 2=π ( 1 .3×10−3

2 )2

=1. 33×10−6m2

The resistance,

R=ρ

LA

=( 9.7×10−8 )( 0 .50

1.33×10−6 )=0 .037Ω

4.4 Series and parallel resistors

- In an electric circuit, resistors can be connected in many ways, depending on their functions in the circuit. Two of the most commonly used ways to connect resistors are by series and parallel connections. Figure 4.3a and 4.3b shows how a group of resistors is connected in series and in parallel.

4.4.1 Resistors in series

- When one or more resistors are wired so that the same current flows through each other, the resistors are said to be wired in series. From Figure 4.3a, the electric current traversing through R1, R2 and R3 is the same current, i.

The potential difference across each resistor is given by Ohm’s law whereV1 = iR1, V2 = iR2 and V3 = iR3.

- From Kirchhoff’s Voltage Law,

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∑ ΔV i= 0 V−V 1−V 2−V 3=0

V−iR1−iR2−iR 3=0 V=iReq

where Req = Equivalent resistance = R1 + R2 + R3.

For n numbers of resistors connected in series, the equivalent resistance is given by

Req=∑i=1

n

R i(4.7)

4.4.2 Resistors in parallel

- When one or more resistors are wired so that the same potential difference occurs across each of the resistors, the resistors are said to be wired parallel to each other. From Figure 4.3b, the electric current supplied by the battery, i1 is divided into i2, i3 and i4 when traversing through each resistor. The magnitude of each current depends on the value of R1, R2 and R3.

- From Kirchhoff’s Current Law,

i1=i2+ i3+i4VReq

=VR1

+VR2

+VR3

VReq

=VR1

+VR2

+VR3

1Req

=1R1

+1R2

+1R3

For n numbers of resistors connected in parallel, the equivalent resistance is given by

1Req

=∑i=1

n1Ri (4.8)

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Figure 4.4

i1

i2

R1

R2R34

Figure 4.5

i1

i2

R1

R234

Figure 4.6


Example 4.4

a) Find the equivalent resistance for the network of resistors in Figure 4.4.

b) Find the current through the resistor R2 if = 0.60 V.

Solution

a) R3 and R4 is parallel to each other

⇒1R34

=1R3

+1R4

R34=R3R4

R3+R4

=1. 0Ω

R34 is shown in Figure 4.5.

R2 and R34 is in series to each other

⇒R234=R2+R34=2. 0Ω

R234 is shown in Figure 4.6.

From Figure 4.6,

⇒Req=

R234 R1

R234+R1

=1 . 0Ω

b) The current through R2 is i2. From Figure 4.6, when i2 flow through R234, the voltage drop across the resistor, V234 = = 0.60 V. The current through R2 is

i2=ξR234

=0. 602. 0

=0 . 30 A

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4.5 Power in electric circuit

- If a charge q moves through a potential difference V across a component, the change in electric potential energy of the charge is

UE = qV (4.9)

The rate at which the energy conversion is the power, P. From definition,

P=ΔU E

Δt=Δ (qV )Δt

=V ΔqΔt

=iV (4 . 10 )

Equation (4.10) is the equation for the electrical power supplied by the battery to the circuit components.

- Electric current flows in a resistor when an emf gives rise to a potential difference between one end and the other. When an electric current flows passes through a resistor, some of the electrical energy is converted into thermal energy due to the resistance of the conductor and is lost or dissipated from the resistor.

From the definition of resistance, the potential drop across a resistor is

V = iR

The rate at which energy is dissipated in a resistor can be written as

P=iV =i (iR)=i2R (4 . 11) or

P=iV=(VR )V =(V )2

R( 4 . 12)

Equation (4.11) and (4.12) are known as the Joule’s law.

Example 4.5

A flashlight is powered by two batteries in series. Each has an emf of 1.50 V and an internal resistance of 0.10 . The batteries are connected to the light bulbs by wires of total resistance 0.40 . At normal operating temperature, the resistance of the filament is 9.70 .

a) Calculate the power dissipated by the bulb.b) Calculate the power dissipated by the wires and the net power supplied by the batteries.

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Figure 4.7


Solution

a) All circuit elements are in series. The circuit of the flashlight is shown in Figure 4.7.

Equivalent resistance of the circuit,

Req=R f+Rw+2r int

=9 .70Ω+0. 40Ω+2 (0 .10Ω)=10 .30Ω

The electric current flowing in the circuit,

i=ξ totalReq

= 3 . 0010.30

=0 .29 A

The power dissipated from the light bulbs,

Pf=i2R f=(0 . 29 )2 (9 .70 )=0 . 82 W

b) The power dissipated by the wires, Pw=i2Rw=(0 .29 )2 (0. 40 )=0 .034 W

The net power supplied by the batteries,

Pnet=P total−Pint=iξtotal−i2rint

=(0 . 29 ) (3 . 00 )− (0. 29 )2 (0 .1+0 .1 )=0 . 857 W

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Figure 1


Supplementary Problems

A) Electric current, battery and emf

1. The starter motor in a car draws 220.0 A of current from a 12.0 V battery for 1.20 s. a) How much charge is pumped by the battery?b) How much electrical energy is supplied by the battery?c) What is the power supplied by the battery?(Answer : 264 C, 3.17 kJ, 2.64 kW)

2. A certain car battery with a 12 V emf has an initial charge of 120 A.h. Assuming the the potential difference across the battery terminals stays constant until the battery is completely discharged, for how long can it deliver energy at a rate of 100 W?(Answer : 14 hrs 24 min)

3. A battery delivering a current of 55.0 A to a circuit has a terminal voltage of 23.4 V. The electrical power being dissipated by the internal resistance of the battery is 34.0 W. Find the emf of the battery.(Answer : 24.0 V)

4. i) A storage battery of emf 6.4 V and internal resistance 0.08 , is being charged by a current of 15 A. Calculate a) the power loss in internal heating of the battery. b) the rate at which energy is stored in the battery

ii) A battery with a 6.00 V emf and an internal resistance of 0.60 supplies current of 1.20 A . a) What is the voltage across its terminals?

b) What is the net power supplied by the battery?(Answer : i) a) 18 W b) 96 W , ii)a) 5.28 V b) 6.34 W)

5. A car battery with a 12 V emf and an internal resistance of 0.040 is being charged with a current of 50 A. a) What is the potential difference across the terminals of the battery?b) At what rate is the energy being dissipated as thermal energy from the battery? (Answer : 10 V, 100W)

B) Resistance and resistivity, Series and parallel resistors, Power in electric circuit

6. A cylindrical copper cable carries a current of 1200 A. There is a potential difference of 0.016 V between two points on the cable that are 0.24 m apart. What is the radius of the cable? (Cu

= 1.72 x 10-8 m)(Answer : 0.01 m)

7. Determine the equivalent resistance between the points A and B for the resistive circuit shown in Figure 1.

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Figure 2

12Ω

24Ω

15Ω

276V

A

B

Figure 3

2.0Ω 1.0Ω

4.0Ω

R

Figure 4

Figure 5

50Ω

90Ω

20Ω

40Ω

20Ω

120 V

a b


(Answer : 4.6 )8. The circuit in Figure 2 contains five identical resistors.

The 45 V battery delivers 58 W of power to the circuit.a) What is resistance of each resistor?b) Determine the potential difference across the R3

resistor. (Answer : a) 25 b) 6.45 V)

9. From the circuit in Figure 3,a) what is the resistance between terminals A and B?b) calculate the total electric current in the circuit.c) a 276 V emf is connected to the terminals A and B.

What is the current in the 12 resistor?(Answer : a) 23 b) 12 A c) 8 A)

10. From the circuit in Figure 4, if =93.5 V, and the current in the 4 resistor is 17 A, what is the value of the unknown resistor R?

(Answer : 5)

11. Consider the circuit in Figure 5.a) What current flows from the battery?b) What is the potential difference

between points a and b?c) Determine the power dissipated in the

50.0 resistor.(Answer : a) 1.1 A, b) 39.6 V, c) 60.5 W)

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chapter 4

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