Chapter 3: Electric Current and Direct-Current Circuit

Chapter 3: Electric Current and Direct-Current Circuit n this chapter, we are going to discuss both the microscopic aspect and macroscopic aspect of electric current. Direct-current is current that flows in one direction.

Overview Electric Current Conduction of Electricity dq dt Ohm s Law Electrical Energy W t Drift elocity esistivity Electrical Power v d nae A l P Dependence of esistance on Temperature T T 0 0

Overview Direct-Current Circuit Electromotive Force, nternal esistor & Potential Difference terminal r First Law Kirchhoff s Laws Second Law Potential Divider 2 in out Potentiometer 2 l l 2 esistors Wheatstone Bridge n Series n Parallel X 2 3

3. Electric Conduction Describe microscopic model of current Define and use electric current, dq dt Learning Objectives

Mechanism of Electric Conduction in Metals v d 0 Before applying electric field Electron move freely and random. Frequently interact with each other. Drift velocity is zero because the free electrons are in constant random motion.

Mechanism of Electric Conduction in Metals After applying electric field E v d v d nae F e The freely moving electron experience an electric force and tend to drift towards a direction opposite to the direction of electric field (positive terminal of the battery). The electric current is flowing in the opposite direction of the electron flows. Drift velocity is the mean velocity of the electrons parallel to the direction of the electric field when a potential difference is applied.

Electric Current Electric current, is defined as the total charge, Q flowing through an area per unit time, t. (O the rate of charge flow through a conductor) Mathematically, Q t O dq dt t is a scalar quantity. The S.. unit for electric current is ampere (A).

Example A current of 2.0 A flows through a copper wire. Calculate a. the amount of charge, and b. the number of electrons flow through a cross-sectional area of the copper wire in 30 s. (Given the charge of electron, e =.60x0-9 C)

Example Solution

3.2 Ohm s Law and esistivity State and use Ohm s Law Define and use resistivity, A l Learning Objectives

Ohm s Law Ohm s law states that the potential difference across a conductor, is directly proportional to the current, through it, if its physical conditions and the temperature are constant. where T is constant Mathematically, constant Ohm s Law equation

Ohm s Law Graphically, Must start from zero Not all materials obey Ohm s law. Materials that obey Ohm s law are materials that have constant resistance over a wide range of voltage. These materials are called ohmic conductor. Examples: pure metals

esistivity esistivity is defined as the resistance of a unit crosssectional area per unit length of the material. t is a measure of a material s ability to oppose the flow of an electric current. A l where l : length A : cross-sectional area t is a scalar quantity Unit is ohm meter ( m) esistivity depends on the material. Same materials have same resistivity. t only changes when the temperature of wire/ material changes.

Example 2 A wire 4.00 m long and 6.00 mm in diameter has a resistance of 5 m. A potential difference of 23.0 is applied between both end. Determine a. the current in the wire. b. the resistivity of the wire material.

Example 2 Solution

Example 3 Two wires P and Q with circular cross section are made of the same metal and have equal length. f the resistance of wire P is three times greater than that of wire Q, determine the ratio of their diameters.

Example 3 Solution

3.3 ariation of esistance with Temperature Explain the effect of temperature on electrical resistance in metals. Use resistance, T T 0 0 Learning Objectives

Electric esistance of A Metal The resistance of a metal (conductor) depends on a. the nature of the material, (, resistivity) b. the size of the conductor, (l, the length and A, cross-sectional area) l A c. the temperature of the conductor. The resistance of metals increases with increasing temperature. (T, )

The Effect of Temperature on Electrical esistance in Metals Explanation:. As temperature increases, the ions of the conductor vibrate with greater amplitude. 2. More collisions occur between free electrons and ions. 3. These electrons are slowed down thus increases the resistance.

ariation of esistance with Temperature The fractional change in resistance per unit rise in temperature is known as temperature coefficient of resistance, α: α is a constant value and it is depends on the material. T 0 T T T T T T 0 0 0 0 0 0 0 0

ariation of esistance with Temperature The resistance of a metal can be represented by the equation below: T T 0 0 where = the resistance at temperature T, o = the resistance at temperature T 0 = 20 o C or 0 o C, T = final temperature T o = reference temperature (20 o C or 0 o C) = the temperature coefficient of resistance ( o C - )

Example 4 A platinum wire has a resistance of 0.5 Ω at 0 C. t is placed in a water bath where its resistance rises to a final value of 0.6 Ω. What is the temperature of the bath? (Given the temperature coefficient of resistance of platinum is 3.93 0-3 C - )

Example 4 Solution

Example 5 A copper wire has a resistance of 25 mω at 20 C. When the wire is carrying a current, heat produced by the current causes the temperature of the wire to increase by 27 C. a. Calculate the change in the wire s resistance. b. f its original current was 0.0 ma and the potential difference across wire remains constant, determine the final current of the copper wire. (Given the temperature coefficient of resistance of copper is 6.80 0-3 C - )

Example 5 Solution

3.4 Electromotive Force (emf), nternal esistance and Potential Difference Define emf, ε of a battery Explain the relationship between emf of a battery and potential difference across the battery terminals Use terminal voltage r Learning Objectives

What is inside a battery

How do batteries work?

Electromotive Force, ε ε Electromotive force (e.m.f.), is defined as the energy provided by the source (battery/ cell) to each unit charge that flows from the source. t is the maximum potential difference across its terminals when it is not connected to a circuits. Unit is volt ().

Terminal Potential Difference, Terminal potential difference (voltage), is defined as the work done in bringing a unit (test) charge from the negative to the positive terminals of the battery through the external resistance only. t is the potential difference across the terminals of a battery when there is a current flowing through it. Unit is volt ().

nternal esistance, r n a cell or battery, the negative ions are attracted by anode and the positive ions are attracted by the cathode. The flow of these ions produces current. However the collisions between the ions and the recombination of opposite ions reduce the flow of current. This resistance in the cell is called internal resistance, r.

elationship between ε, and r terminal lost r r Total resistance in the circuit

Note terminal < ε when the battery of emf ε is connected to the external circuit with resistance. terminal > ε when the battery of emf ε is being charged by other battery. terminal = ε when the battery of emf ε has no internal resistance (r = 0) and connected to the external circuit with resistance.

Example 6 A battery of internal resistance 0.3 is connected across a 5.0 resistor. The terminal potential difference measured by the voltmeter is 2.5. Calculate the e.m.f. of the battery.

Example 6 Solution

Example 7 When a 0 resistor is connected across the terminals of a cell of e.m.f. and internal resistance r, a current of 0.0 A flows through the resistor. f the 0 resistor is replaced with a 3.0 resistor, the current increases to 0.24 A. Find and r.

Example 7 Solution

Example 8 A battery has an e.m.f. of 9.0 and an internal resistance of 6.0. Determine a. the potential difference across its terminals when it is supplying a current of 0.50 A, b. the maximum current which the battery could supply.

Example 8 Solution

3.5 Electrical Energy and Power Use power, P = electrical energy, W = t Learning Objectives

Electrical Energy, W Electrical (potential) energy, W is the energy gained by the charge Q from a voltage source (battery) having a terminal voltage. The faster the electric charges are moving the more electrical energy they carry. The work done by the source on the charge is given by: Given Q = t, W Q W t For source

Electrical Energy, W Since =, electrical energy can also be written as: W 2 t O W 2 t The unit for electrical energy is Joule (J).

Electrical Power, P Electric power is defined as the rate of energy delivered to the external circuit by the battery. P W t t t P For source Since =, electrical power can also be written as: P 2 O P 2 The unit for electrical power is Watt (W).

Example 9 a. Calculate the resistance of a 40 W automobile headlight designed for 2? b. The current through a refrigerator of resistance 2 Ω is 3 A. What is the power consumed by the refrigerator?

Example 9 Solution

Example 0 n figure above, a battery with an e.m.f. of 2 and an internal resistance of.0 is connected to a 5Ω resistor. Determine a. the rate of energy transferred to electrical energy in the battery, b. the rate of heat dissipated in the battery, c. the amount of heat loss in the 5.0 resistor if the current flows through it for 20 minutes.

Example 0 Solution

evision a. A dry cell of e.m.f..50 is connected in series with a resistor and another battery. When a current of 3.0 A flows from the cell, the voltage across the cell is 0.42. What is the internal resistance of the dry cell? (0.36 Ω) b. The e.m.f. of a cell is.5 and its internal resistance is 0.5 Ω. Determine the power supplied to an external resistor of resistance 2.5 Ω. (0.625 W)

3.6 esistors in Series and Parallel Derive and determine effective resistance of resistors in series and parallel Learning Objectives

esistor in Series The symbol of resistor in electrical circuit can be shown in esistors in Series O 2 3 2 3

esistor in Series The same current flows through each resistor where 2 3 Assuming that the connecting wires have no resistance, the total potential difference, is given by 2 3 From the definition of resistance, ; 2 2 ; 3 3 ; eq Therefore, eq 2 3 eq 2 3

esistor in Parallel esistors in Parallel 3 3 3 3 2 2 2 2 2 3 3 2

esistor in Parallel There is the same potential difference, across each resistor where 2 3 Charge is conserved, therefore the total current in the circuit is given by 2 3 From the definition of resistance, ; 2 ; 3 ; 2 3 Therefore, eq 2 3 eq eq 2 3

Example For the circuit shown below, Calculate : a. the total resistance of the circuit. b. the total current in the circuit. c. the potential difference across 4.0 resistor.

Example Solution

Example 2 For the circuit shown below, calculate the equivalent resistance between points x and y.

Example 2 Solution 3 4 2 4 2 2 5 5 3 3

A 3 4 2 3 2 3 2 4 5 B 5 Since 4 and 5 are in series: X 4 5 X 4 5 3 is constant from point A to point B. A 3 Step 2 2 2 X X 3 3 B

A 3 2 3 2 3 2 X B X Since X and 2 have same potential difference (point A and B), they are in parallel: Y Y Y X X 2 3 2 2 Step 2 Y Y Y 3 3

A Y Y 3 Y Since Y and 3 are in series: Z Z Y 3 Y is constant from point A to point C. Y 3 C 3 A Step 3 Z Z Y C

A C Z Z Y Since Z and have same potential difference (point A and C), they are in parallel: eq eq eq Z Y Z Step 4 eq eq eq Answer: eq 0.79

3.7 Kirchhoff s Laws State and use Kirchhoff s Laws (Maximum two closed circuit loops) Learning Objectives

Kirchhoff s First Laws Also known as Kirchhoff s Junction/Current Law States the algebraic sum of the currents entering any junctions in a circuit must equal the algebraic sum of the currents leaving that junction. Obeys the principle of conservation of charge. Mathematically, For example: in out

Kirchhoff s Second Laws Also known as Kirchhoff s Loop/oltage Law States in any closed loop, the algebraic sum of e.m.f.s is equal to the algebraic sum of the products of current and resistance or in any closed loop. Obeys the principle of conservation of energy. Mathematically,

Kirchhoff s Second Laws Sign convention for e.m.f., ε: Sign convention for the product of :

Problem Solving Strategy For ONE closed circuit. Label the diagram 2. Apply Kirchhoff s Second Law equation 3. Solve the equations Only one loop, no junction in the circuit: Don t have to apply Kirchhoff s first law ɛ ɛ 2 2 Kirchhoff s second law: 2 2 3 3

Problem Solving Strategy For TWO closed circuits. Label the diagram 2. Apply Kirchhoff s First Law equation 3. Apply Kirchhoff s Second Law 2 equations 4. Use scientific calculator to solve the simultaneous equations ε Kirchhoff s first law: in out ε 2 3 2 2 3 2 3 2 Kirchhoff s second law: Loop : 2 22 Loop 2: 2 22 3 ( ) 3

Example 3 Using Kirchhoff s rules, find the current in each resistor.

Example 3 Solution

Example 4 Calculate, ε X and

Example 4 Solution Step : Label the diagram ε X = 7.2 4.0 Ω 0.84 A ε Y 2 8.0 Ω 2 0.32 A Step 2: Apply Kirchhoff s First Law in 2 out 2 0.84 0.32.6A

Example 4 Solution Step 3: Apply Kirchhoff s Second Law ε X = 7.2 ε Y 4.0 Ω 2 8.0 Ω 0.84 A Loop : X Y 2 2 0.32 A 2 Loop 2: Y 2 2 2

Example 4 Solution Step 4: Solving the simultaneous equations Loop : Y X Y 7.2 2 0.844.0.68.0 Y 9.84 Loop 2: Y 9.84 Y Y (.68.0) (0.32 ) 33.0

Example 5 Using Kirchhoff s rules, find the current in each resistor. Answer: =3.75 A, 2 = 5.0 A, 3 =.25 A

Example 6 For the circuit shown below, Given = 8, 2 = 2, 3 = 3, = and = 3 A. gnore the internal resistance in each battery. Calculate a. the currents and 2. b. the e.m.f. 2. Answer: A, 4 A, 7

Example 7 For the circuit shown below, Determine a. the currents, 2 and 3, b. the potential difference across the 6.7 resistor, c. the power dissipated from the.2 resistor. Answer: =0.72 A, 2 =.03 A, 3 =.75 A ; 6.90 ; 3.68 W

3.8 Potential Divider Explain principle of potential divider Use equation of potential divider, 2 Learning Objectives

Potential Divider Also known as voltage divider. A potential divider produces an output voltage that is a fraction of the supply voltage. This is done by connecting two resistors in series as shown in figure below. 2 Since the current flowing through each resistor is the same, thus eq and eq 2 2 2

Potential Divider Therefore, the potential difference (voltage) across is given by 2 Similarly, 2 2 2 2 2

Potential Divider esistance and 2 can be replaced by a uniform homogeneous wire as shown in figure below. a l c l 2 2 b Since ρl A Therefore, the potential difference (voltage) across the wire with length l l l 2 and l2 2 l l 2

Summary 2 2 a l c l 2 2 b l eff 2 2 l total l l 2 2

Example 8 esistors of 3.0 Ω and 6.0 Ω are connected in series to a 2.0 battery of negligible internal resistance. Determine the potential difference across the a. 3.0 Ω resistor b. 6.0 Ω resistors

Example 8 Solution

3.9 Potentiometer and Wheatstone Bridge Explain principles of potentiometer and Wheatstone Bridge and their applications. Use related equations for potentiometer, and for Wheatstone Bridge, 2 l l 2 2 X Learning Objectives

Potentiometer 0 0 G No current flow 8 0

Potentiometer The working of potentiometer is based upon the fact that fall of the potential across any portion of the wire is directly proportional to the length of the wire provided wire has uniform cross section area and constant current flowing through it. A + - x G C (Unknown oltage) l because l B Jockey The potentiometer is balanced when the jockey (sliding contact) is at such a position on wire AB that there is no current through the galvanometer. Galvanometer reading = 0

Application : To measure the unknown e.m.f. C A B + - x (Unknown oltage) G Jockey The potentiometer is balanced when the reading of galvanometer is equal to zero (no current through the galvanometer). X AC

Application : To measure the unknown e.m.f. A l Total G C B a l c l 2 2 b AC BC Same as potential divider l Total l AC AC l BC BC

Application 2: To compare two e.m.f. l l l A earranging: l () A

Application 2: To compare two e.m.f. 2 l2 l2 l A 2 earranging: 2 l (2) A 2

Application 2: To compare two e.m.f. () (2) = A l A l 2 2 2 2 l l or 2 2 l

Application 3: To measure the internal resistance of a cell l0 l0 earranging: l A 0 l () A 0

Application 3: To measure the internal resistance of a cell Circuit t l l l A Circuit 2 earranging: t l A (2)

Application 3: To measure the internal resistance of a cell When switch is CLOSED, circuit 2 has current flow now. When there is current flow, the battery has internal resistance due to the chemical reaction inside the battery. So for circuit 2, t r earranging r t t t t Substitute r (3)

Application 3: To measure the internal resistance of a cell Substitute () and (2) into (3): earranging: l l l r 0 c x m y 0 l 0 l r l or l l r 0

Example 9 Consider a potentiometer: f a standard battery with an e.m.f. of.086 is used in the circuit, the galvanometer reads zero when the resistance is 36 Ω. f the standard battery is replaced by an unknown e.m.f. the galvanometer reads zero when the resistance is adjusted to 48 Ω. What is the value of the unknown e.m.f.?

Example 9 Solution

Example 20 n the potentiometer circuit shown below, PQ is a uniform wire of length.0 m and resistance 0.0. is an accumulator of e.m.f. 2.0 and negligible internal resistance. is a 5 resistor and 2 is a 5.0 resistor when S and S 2 open, galvanometer G is balanced when QT is 62.5 cm. When both S and S 2 are closed, the balance length is 0.0 cm. Calculate a. the e.m.f. of cell 2. b. the internal resistance of cell 2.

Example 20 Solution

Wheatstone Bridge t is used to measured the unknown resistance of the resistor. Figure below shows the Wheatstone bridge circuit consists of a cell of e.m.f. (accumulator), a galvanometer, know resistances (, 2 and 3 ) and unknown resistance x. The Wheatstone bridge is said to be balanced when no current flows through the galvanometer.

Wheatstone Bridge When the Wheatstone Bridge is balanced: AC AD CB DB 2 Potential at C = Potential at D AC AD and BC BD Since =, and 2 2X 23 Dividing gives: 2 3 2 3 2 2 X X

Application: Metre Bridge The metre bridge is balanced when the jockey J is at such a position on wire AB that there is no current through the galvanometer. X l l 2

Example 2 An unknown length of platinum wire 0.920 mm in diameter is placed as the unknown resistance in a Wheatstone bridge as shown in figure below. esistors and 2 have resistance of 38.0 and 46.0 respectively. Balance is achieved when the switch closed and 3 is 3.48. Find the length of the platinum wire if its resistivity is 0.6 x 0-8 m.

Example 2 Solution

Summary Electric Current Conduction of Electricity dq dt Ohm s Law Electrical Energy W t Drift elocity esistivity Electrical Power v d nae A l P Dependence of esistance on Temperature T T 0 0

Summary Direct-Current Circuit Electromotive Force, nternal esistor & Potential Difference terminal r First Law Kirchhoff s Laws Second Law Potential Divider 2 in out Potentiometer 2 l l 2 esistors Wheatstone Bridge n Series n Parallel X 2 3