1 Explain what is meant by electric current.

Current is the rate of flow of electric charge through a given cross-section of the conductor'

2 Define charge and the coulomb.

Charye is lhe properly of some elementary patticles that causes them to exerT electric forceon one another- The charge that passes through a given point is the ptoduct of the steadycurrentflowing past the point and the time during which the current flows.The coulomb is the quantity of electric charye that passes through a given point in a circuitwhen a steady current of one ampere flows for one second,

3 (a) A cuffent of '12 A flows for 20 minutes into an electric cooker. How much charge passedthrough the cooker? [14400 C]

O = tt = (1 2)(20^6q = ua00 c

(b) If 18400 C of charge flows through an air conditioning unit every hour, what current does it

draw? [5.11 A]

t= Q/t = 1u00/(60x60) = 5.11 A

(c) A current of 3 A flows into a television set. For how iong would it take 1500 C of charge to flowthrough it? 18 min 20 sl

t. qI = 1500/ 3 = 500 s. I min 20 s

4 A beam of electron is emitted from a heated filament, resulting in 0.01 C of charge being releasedin 20 s. (Given that the charge on a single electron, e = 1.6 x 10 re c)(a) What is the current? [5 x 10 4 A]

t.0.01/20=5x1o1A

(b) What is the rate of electrons emission? [3.13 x 1ors]

Q = I t= Ne where N is the number of electrons and e is the charge of a single electran.Therefore, Mt = l/e= 5 x 101 A / 1-6 x 1o"e c = 3.13 x 1015 per second.

I

lN4 J_

{

.,-\ _ T j-L! - J_L

Ne=ar0=-rt

A high potential difference is applied between the electrodes of a hydrogen discharge tube so thatthe gas is ionlzed. Electrons then move towards the positive electrode and protons towards thenegative electrode as shown ln Fig. 5.1. ln each second, 5 x 1ors electrons and 2 x lOrB protonspass a cross-section of the tube. What is the currenl flowing in the discharge tube? [1. 12 A]

Fig.5.l

The current due to the flow of protons. nee/ t - (2fl014)0.6x1o4e)/ 1 = 0.32 A

The current due to the flow of electrons = n.e/ t = (5x1O13)(1.6x1,aey 1 = O.8O A

An electron charge -1,6x101e C moving to the teft is equivalent to a positive charge of1.6x101s moving to the right,

The total current I = 0.32 + 0.80= 1.12 A

DeIine potential difference and volt.

The potential difference between two points in a circuit is the amount of electrical energydissipated when a unit charge flows from one point to the other. lt's Sl unit is volt (v).

The volt is the pd. between two points in a circuit if one joule of energy is drssrpafedbetween the two points when one coulomb ot charge flows,

The potential difference between point P and point Q is '15 V and the current flowing through is

0.025 A. During a period of time t, the energy of the charge carriers changes by 6 J when movingfrom P to O. Determine the valLre of time f. [16 s]

Since

(a) Define resistance and ohm

The resistance of a conductor is the ratio of the potential dilference across theconductar to the current flowing through the conductor.It is measured in ohm (!).

The ohm is the resistance of a conductor through which a current of one ampere flowswhen a potential difference of one volt is maintained across it.

t=16 s0.025 r f

(b) Calculate the resistance per metre of a1.7x 10 3om.[8.66x10'?()m 1]

p 1.7x10-s

J

copper wire of diameter 0.500 mm and resistivity

= 8.66 x 10-2 c)m 1

,,Wo

R= p =

* (o.soo, r o '/zr@lzf

Figure 9.1 shows a rectangular block with dimensions -x, 2x, 3r .

2t

Fig.9.1

Electrical contact to the block can be made between opposite pairs of faces (for example, betweenthe faces labeled P). Between which h]Vo faces would the maximum electrical resistance beobtained? [Q]

p(3x)x(2x)

Maximam electrical resistance would be obtained between the faces labeled Q.

10 The resistance of a thermistor decreases significantly as its temperature increases. Sketch al-V gtaph to show how current / in the thermistor depends upon the potential difference I/across it.

Explain why heat is produced in a cuffent carrying conductor.

when current is flowing through the conductor, free electrons collide with the laftice ions.As a rcsult" their kinetic energy is being transferred to the lattice ions, The ions vibrate withincreased amplitude, and cause heat to be produced.

-1(e)2l.x )

,._oi __ox _1fol R"= p/"'- A-ixtzx\ 6\x) "o-A-

^-- er ptzx) -l( e \'' / 3x(x) 3tx'

'11

12 A potentjal difference ydrives a current / through(a) the filament of a torch bulb.(b) a piece of lntrinsic semiconductor.

Sketch graphs, one for each case, to show howvalue which causes significant heating.

(a) the filament of a torch bulb. (b)

, depends on Vas yis increased from zero to

A piece of intrinsic semiconductor,

V

As the temperature gf filamenl rises, the amplitude of vibration of the meta! ions in thelattice increases, The drifting electrons will collide more frequently with the metal ionscausing the resistance of the filament to increase- Thus, the value of V/l increases as thecurrent increases. The graph of I against V becomes /ess and ress steep as the currentincreases from zero,

lntrinsic semiconductot is essentially pure such that the impurities in it do not appreciablyaffect its electrical behavior unless it is heated up. The electric resrslarce of a typicalintrinsic (non doped) semiconductor decreases exponentially with the temperature: theincrease in thermal energy of the valence electrons due to the temperature rise enablesmore of them to break the covalent bonds and become free electrons, Thus more electron-hole pairc are produced which can act as carriers of current, At the same time, there is alsoan increase in rate of atomic vibralion. However, the decrease in resistance due to theincrease in no. of charge carriers predominates fhe rncrease in resistance caused by theincrease in rate of atomic vibration- Hence, the overall effect is that R decleases as y

13 An electrical heating element is to be made from nichrome ribbon 1.0 mm wide and 0.05 mm thick.The element is designed so that the power dissipated will be 750 W when connected ta a 244 Vsupply.(a) Calculate the resistance of the wire required. [76.8 o]

p=L = n= /' =240' -'/6.8 aRP75O(b) Calculate the length of the ribbon required. [3.49 m]

(resistivity of nichrorne - 1.1 x 10 6 o m)

n t ,r,r t?o.Rtl(l lo 0.05 10 I1q J-! > I -'"' -Ll- - r.4o m4 p Ll.l0'

14 A water heater is mafted 230 V, 3000 W. lt is switched on for 5000 seconds For this heater,calculate(a) the current through the heater. ['13.0 A]

the currcnt through the heater, I = PN 2 3000/230 = 13'O A

(b) the resistance of ihe heater. [17.7 O]

the resistance of the heater, R = U/l =230/13.0 = 17.7 o

(c) the energy supplied by the heater dudng this time. [1.5 x 10' J]

the eneryy supplied by the heater during this time E = Pt= 3OO0(5OOO) = 1.5 x 107 J

15 A 240 ymotor runs an electric power drill. Calculate(a) the power when the drill runs freely and the current is 0.2 A. [48 W]

Power, P - lV = 0.2(240) = 48 W

(b) the power when the ddll is ddiling a hole in a piece of wood and the current is 2 A. [480 W]

P=2(240)=48ow

(c) the efficiencywhen drilling wood, jf thedrill bit does 1B00 J of workperminute. 16.25%l

Power output = 1800/60 = s0 # Wetficiency when drilling wood = Pouy' Pi" X'l00ok= 301480 x lOOo = 6.25%

'16 A wire of length 2.50 m and diameter 1.60 mm carries a current of 1.25 A. The wire dissipatesenergy at the rate of 0.810 W. Calculate the resistivity of the material under the above condjtions.

[4.17x107om]

(0.518)[z(1.60 10'+2)'l-,--]-- r-4.17.1r;'om

2.50

17 Define e.m.f. in terms of energy. State and define the S.l. unjt of e.m.f. What is the function of "thesource of e.m.f."?

fhe e.m.f- of a source is defined as the energy transferred by a source in driving unitcharge round a complete circuit.The s"t unit of e.m.f- is VolL The e,m"f" is said to be one volt when it delivers electricalenergy of one joule in driving one coulomb of charge round a complete circuit'A source of e.m,f, is a device capable of transforming other forms of eneryy into theelectfical form so as to dtive a unil charge around the circuit,

P-tR -> n j. n*lo-o.stroI' 1)5'

nl RAR=!- :::., o=AI

18 A cell is connected in series with a 2 c) resistor and a switched as shown in Figure 18.1. Avoltmeter connected across the cell reads 12 V when the switch is open but I V when it is closed.What is the internal resistance of the cell? ['1 O]

Fig. l8.1

Let internal resistance be r,E -12 U V4 tR .- t. U/R - 8/2 = 4 A.Ee l(R + t) -t r!E/l -R212/1-2-1a

(a) State the equation relating curent /, charge Q and time Ior I ='1Q

dt

(b) There is a current of 6.0 A through a component for 200 s. Calculate(i) the charge which flows past a point in the component during this time. [1200 C]

e = tt = 6.0 (200) =1 200 c

(ii) the number of electrons whish pass the point during thrs time. 17.5 x 10211

Q = ne 1 n - qb = 1 2oo/(1.6 x 1 o1s) =7.5 xI 0'?1

19

for steady current , Q: lt

(c)

12V

Fig. 19.1

The circuit of Figure 19.'1 is usedcharacteristic is shown in Figure 19.2,

Fig.19.2

produce the lV characteristic of the lamp. This

(i) A voltmeter and ammeter are required in order to obtain the necessary readings. CopyFig. 19.1, and show the two rneters in appropriate positions.

(ii) For the lamp operating under normal conditions with a potential difference of 12.0 Vacross lt, calculatef. ihe resistance of the larrp. [4.0 O]

From the graph, when V= 12.0 V, l= 3.0 AThus R = U/l c 12.0/ 3.0 = 4.0 a

2. the power supplied to the lamp. [36 W]

Power suqqlied = lV = 3(12.0) : 36 W

(iii) Calculate the resistance of the lamp when the potential difference across it is only6.0 v. [2.5 o]

From the gtaph, when V = 6.0 , I = 2,4 AResistance R. u/l . 6.0/2.4 - 2.5 a

(iv) Explain in microscopic terms why the resistance in (iii) is less than resistance in (ii).

when current is smaller, the frequency of collision of the electrons with the lafticeions is lower. As temperature of the filament is lower, the vibration amplitude of theions will also be smaller, and will not result in fu.ther collisions with the driflingelectrons. Thus the resistance decreases,

(v) Sketch a graph showing how the resistance of the lamp varies with the potentialdifference across it.

From V= 0V to v = 2v, the graph is linear - resistance (V/l) is constantAftet V= 2 V, the gradient becomes gentler ) resistance Ml) increasesR=Vl=41.4=1.4O

VN

(a)20

(b)

Use energy considerations to distinguish between the terrns potential difference andelectromotive force.

Pd- is used when electrical energy is being converted into other forms of energy when aunit charge passes frcm one point to another, wheteas e-m-f. is used when electricalenergy is being produced from other foms of energy to drive charges round the circuit.

using the definition of potential difference, show that P, the power dissipation in a resistor of aresistance R is given by

,=tR

where y is the potential difference across the resistor-

By definition of potential difference,

where t is the time and I is lhe current.

'r''2.t p = tv = (:)v = i Ghown)

W W/t P

Q Q/t I

(c) A battery of e.m.f. E and internal resistance r is connected in series with a resistor ofresistance R as shown in Fig.20.

Give expressions, in terms of E, r and R for1. the current / in the circuit,

E=l(R+r)

2- the power P,? dissipated in the externa' resistor

Power dissiDated- P.= fR = ! r;tR+l'

(i)

(ii) The battery generates total power Pn Show that the fraction fi is given by

PpRe tR-f

Pr= l'z(R+t)

P. = l'R =:+'' (R*4'

PN I2R R

+= 76-n= B-i

(d) A car batiery of e.m.f. 12 V and internal resistance 0.014 O delivers a current of 1'10 A whenfirst connected to the stafter motor.

(i) CalculateL the resistance of the starter motor, [0.0951 Q]

E.t(R+r)12 = 11o (R + 0.014)R = 0.0951 A

2. the fraction of the total power which is dissipated in the battery. [0.128]

Frcction of totat power dissipated in the battery = ZP, r 0"014

-

o 1?BPt (R+r) 0-0951+0.014 -

(ii) After prolonged use, the intemal resistance of the battery may increase. State and explainhow the performance of the battery is affected by the increase in internal resistance.

The performance of the battery correspond to the powet PR of the motor given bythe equation from c(i),

An increase in the intemal resistance wiu hence reduce the power of the motor,reducing the peiormance of the batlery according to the equation.

*** End ofTutorial ***