Bab 4

CURRENT & RESISTANCE

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C U R R E N T AND R E S I S T A N C E

4.1 ELECTRIC CURRENT

In the previous chapter we have been studying static electricity that is electric charges at rest. In this chapter we begin our study of charges in motion as an electric current.

The electric current in a wire is defined as the net amount of charge that passes through it per unit time at any point. Thus, the average current I is defined as

I =

Where ΔQ is the amount of charge that passes through the conductor at any location during the time interval Δt.

Example 4.1

A steady current of 3.5A flows in a wire for 20.0 minutes. (i) How many charges passed through any point in circuit? (ii) How many electrons would this be?

Solution

(i) = (3.5)(20X60) = 4200 C

(ii) Number of electrons = 4200 X 6.242 x 1018 = 2.62 X 1022

Exercise 4.1

(a) How many electrons per second pass through a section of wire carrying a current of 0.70 A?

(b) What is the current through an 8.0- toaster when it is operating on 120 V?

PHY 193 Physics For Engineering II UiTM Pulau Pinang

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The direction of conventional current flow is that of positive charge. In a wire, it is actually negatively charged electrons that move, so they flow in direction opposite to the conventional current. Positive conventional current always flows from a high potential to a low potential as shown in Figure 4.1.

Figure 4.1: The Flow of electric current

4.2 OHM’S LAW AND RESISTIVITY

Ohm’s law states that the current in a good conductor is proportional to the potential difference applied to its ends. The proportionality constant is called resistance R of material, so V = IR. The unit of resistance is the ohm ( Ω ), where 1 Ω = 1 V/A.

The resistance R of a wire is inversely proportional to its cross–sectional area A, and directly proportional to its length L. The general property of the material actually is depending on the resistance and we call as resistivity (). That is resistance is a property of a particular piece of material with a particular size and shape.

R = ρ ( is the Greek letter rho)

Example 4.2


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A nichrome wire is 100cm in length and 0.6 mm in radius. When connected to a potential of 2V, a current of 4A exists in the wire. Find the resistivity of this nichrome wire.

Solution

=

=

=

= 5.65 X 10-7 m

Exercise 4.2

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.00mm. Conductor B is a hallow tube of outside diameter 2.00mm and inside diameter 1.00mm. What is the resistance ratio RA/RB, measured between their ends?

4.3 ELECTRIC POWER

The rate at which energy is transformed (absorbed by) in a resistance R from electric to other forms of energy (such as heat and light) is equal to the product of current and voltage or energy transformed per unit time.

P = = = = IV

and for resistors can be written with the help of Ohm’s law as

P = I2R =

The SI unit for power is called Watt ( W ).


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Example 4.3

A current of 0.5A flow through a 200 Ω resistor. The power lost will appear as heat in resistor. How much powers is lost in resistor.

Solution

P = I2R = 0.52 X 200 = 50 W

Exercise 4.3

A current of 3.0A is passed through a lamp for 2 minutes using a 6V power supply. Find the energy dissipated by this lamp during the 2 minutes.

4.4 RESISTANCE IN COMBINATIONS

Resistors in Series

As in a Figure 4.2 the potential difference across each resistor is found from Ohm’s law:

V1 = IR1 V2 = IR2 V3 = IR3

We want to replace the series of resistors with a single equivalent resistor R eq connected between the same two terminals A and B.

The total potential difference between A and B is V = V1 + V2 + V3

Since the potential difference between A and B across Req also must be V, we have

V = IReq . Using the Ohm’s law, we obtain IReq = IR1 + IR2 + IR3

Req = R1 + R2 + R3

The argument can be extended to any number of resistors in series :

Req = R1 + R2 + R3 + … (resistors in series )


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V3V2V1

BA

VI

R1 R2 R3

Figure 4.2


Resistors in Parallel

In a parallel circuit, Figure 4.3, the total current I that leave the battery break into three branches. We let I1, I2 and I3 be the currents through each of resistor, R1 , R2 , and R3

respectively. The current flowing into a junction must equal the current flowing out of the junction. Thus

I = I1 + I2 + I3

Hence the full voltage of the battery is applied to each resistor so

I1 = , I2 = and I3 =

The equivalent resistance Req must satisfy

I =

We combine the equation above

I = I1 + I2 + I3

= + +

we divide out the V from each term, we have

= + + … (resistors in parallel )

Example 4.4

Five resistors are connected as shown in Figure 4.4. Each has a resistance of 4. What is the equivalent resistance between points A and B?


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R6

I3

I1

I2

R1

R2

R3

-+

V

Figure 4.3

R2

R5R3

R4

R1

A B


Solution

R2 , R3 and R4 (Parallel)

=

R234 =

R234 , R1 and R5 (series)

Req = R234 + R1 + R5 = 1.33 + 4 +4 = 9.33

Exercise 4.4

Six resistors are connected as shown in Figure 4.5. Each has a resistance of 4. What is the equivalent resistance between points A and B?

Example 4.5

Two resistances are connected (a) in series and (b) in parallel, to 12.0V battery as Figure


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Figure 4.4

R1

R5R4

R3R2

B

R1

A

Figure 4.5


4.6. What is the equivalent resistance of each circuit and the current through each resistor? (a)

(b)

Solution

(a) R1 and R2 (series)

Equivalent Resistance Req = R1 + R2 = 12

Current

(b) R1 and R2 (parallel)

Req = 2.67

Current , Current

Current

Exercise 4.5

From the Figure 4.7, find the


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I

I2

I1R1=4

R2=8

V = 12V

I

V = 12V

R2 = 8R1=4

Figure 4.6


(a) current I1 and I3

(b) difference potential at point CD,CF and BG(c) current I4 and I(d) total equivalent resistance.(e) voltage supplied V if internal resistance are negligible.

4.5 EMF AND TERMINAL VOLTAGE

A device such a battery or an electric generator that transforms one type of energy (chemical, mechanical, light and so on ) into electric energy is called a seat or source of electromotive force or emf. The symbol ε is usually used for emf.

Thus, a battery itself has some resistance, which we called its internal resistance ( r )

The two point a and b in a diagram represent the two terminals of the battery. What we measure is a terminal voltage Vab.

Vab = ε ( If no current drawn from a battery)

Vab = ε – Ir ( when a current I flows from a battery)Example 4.6


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Vab

εr

Figure 4.7


A 9.0V battery whose internal resistance r is 0.5Ω is connected in the circuit shown in Figure 4.8.(a) Find the equivalent resistance for external loaded(b) How much current is drawn from the battery?(c) What is the terminal voltage of a battery? (d) What is the current in the 10 and 5Ω resistor?

Solution

(a) R2 and R3 (parallel)

R23 = 2.67

R1 and R23 (series)R123 = R1 + R23 = 7.67

R123 and R4 (parallel)

R1234 = 4.34

R1234 and R5 (series)Req = 4.34 + 6 = 10.34


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C D

BA

R1=5 R3 =

4

R2 = 8

R4 = 10

R5 = 6 = 9 V r=0.5

Figure 4.8


(b)

(c) VT = IReq = 0.83(10.34) = 8.58V

(d) VT = VAB

VCD = VT – V(drop R5) = VT – (IR5) = 8.58 – (0.83)(6) = 3.6V

Exercise 4.6

From the Figure 4.9, find(a) the equivalent resistance between point A and D(b) the current drawn from battery I(c) the terminal voltage(d) current I2 and I3

(e) potential difference at point BE and CF


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Figure 4.9


Tutorial 4.1

1. A service station charges a battery using a current of 5.7 A for 7.0 hour. How much charge passes through the battery?(Answer : 1.4 x 105C )

2. A conducting wire has a 1 mm diameter, a 2.0m length and a 50 m resistance. What is the resistivity of the material?(Answer : 2 X 10-8 m )

3. A heater uses nichrome wire ( = 1 X 10-6 m) and generates 1250 W when connected across a 240 V line. How long must the wire be, if it cross sectional area is 0.2X10-2 cm2 . (Answer : 9.216 m )

4. A cylindrical cooper cable carries a current of 1200A. There is a potential difference of 1.6 x 10-2V between two points on the cable that are 0.24 m apart. What is the radius of the cable? (ρCooper = 1.72 x 10-8 Ω.m)(Answer : 9.9 x 10-3m)

5. An electric heater has a rating of 3 kW at 240 V. Calculate the length of heater element if resistor per unit length is 2 m-1.(Answer : 9.6 m )

6. Two resistors 30 and 20 are connected in parallel and the combination is connected to a source 120V with an internal resistance 1.(a) Calculate the equivalent external resistance in the circuit.(b) Calculate the current through each resistor.(Answer : 12 , I1 = 3.7A , I2 = 5.54 A )

7. Refer to Figure 4.10. When the switch S closed, the voltmeter V shows a reading of 7.5 V and the current, I flowing in the circuit is 0.5 A. If the e.m.f of the battery is 9V, find the internal resistance in the battery, r and the load resistance R. (Assume that the voltmeter is an ideal with high resistance).


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Figure 4.10

R

v

SI


(Answer : r= 3 , R = 15 )

8. In Figure 4.11, R1, R2 and R3 are external load and r is internal resistance. Find

(a) Req for external load.(b) Current I(c) Total power that received by external load(d) If the internal resistance is zero, calculate the current I and total power

received by external load.(e) Calculate the power percentage reduced due to internal resistance.

(Answer : 12 , 0.5625A , 3.8W , 0.75A , 6.75W , 43.7% )

9. Refer to the Figure 4.12, the battery has an e.m.f of = 12V and internal resistance of r = 3.0. Find(a) equivalent resistance for external load(b) current I(c ) potential difference between point a to c (d) the current through R3

(e) potential difference between point b to c(f) the current through R4

(g) the total power dissipated in the R1

(Answer : 6.65, 1.24A, 4.526V, 0.754A, 0.64V, 0.32A, 4.61W)


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R1 = 4

r= 4

R2= 4

R3 = 10

= 9V

Figure 4.11

Ia

R1=3


10. From the Figure 4.13, the e.m.f value is 8.0 V and internal resistance r= 0.5Ω(a) Find the equivalent resistance for external load(b) Find the current i(c) Find the terminal voltage(d) Find the current through the resistor R5

(Answer : 9.46, 0.8A, 7.57V , 0.584A)


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R3=6

R4=2 R5=4

c

b

r

R2=8

Figure 4.12

Figure 4.13


11. From the Figure 4.14 shown, find(a) Req between point AB, BC and AD(b) Current i1, i2 and i3

(c) The electric power received between point AB(d) The electric power supplied by battery(e) The electric power used at point AD(f) The electric power lost due to internal resistance(g) The percentage of electric power that lost due to internal resistance.(Answer : (a) 1 , 2 and 5 (b) 2.3A , 1.15A and 1.53 A (c) 5.29W (d) 36.8W

(e) 26.5W (f) 10.6W (g)28.8%

12. (a) A copper wire has a length of 150 m and a diameter of 1.00 mm. If the wire is connected to a 1.5 V battery, how much current flows through the wire?

(ρCopper = 1.72 x 10-8 Ω.m)

(b) Find the number of electrons that pass a point in a wire carrying 8 A during 10s 1 electron = 1.602 x 10-19 C

APR 2009 (Answer : 0.4563 A, 5 x 1020 electrons)

13. Figure 4.15 shows a circuit consisting of 5 resistors connected to a 24 V battery. If R1 = 10 Ω and R2 = 15 Ω, calculate(a) the equivalent resistance of this circuit(b) the current through R2


Figure 4.14

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R4 = 3 R8 = 2

r = 2


(c) the power dissipated in R2

OCT 2008

(Answer : 36.54 Ω, 0.505 A, 3.83 W)

14. A 115 m long copper wire (resistivity 1.7 x 10-8 Ωm) has a resistance of 8.0 Ω. Calculate the diameter of the wire.

APR 2008 (Answer : 5.57 x 10-4 m)


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Figure 4.15

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