Self"Check Cluestions:1 State the formula for calculating the effective resistance of resistols connected in

(a) se es and

Resistors in series: R=Rr+Rz +.......... + Rn

(b) parallel.

. ...1 1 1 1Resrstors rn oarallel: - |' RRr R2 R,

(c) State the equation for calculating the effective resistance of t\to resistors that areconnected in parallel.

For !{q resislors Ri and R2 in parallel, eflective rer;"t"n"" n = {'Ra,

(d) State the equation for calculating the current flowing through one of a set of parallelresistors using the concept of current divider.

I-=R"tr"'uu"*,

2 Sfale Kirchoff's Firsl Law

Kirchoff's First Law states that at any junction point, the sum of all currents ente ng the junctionmust equal the sum of all currents leaving the junction.

3 State the equation for calculating the potential across one of a series of resistors using thepotential divider principal.

,' R"rrc.u," ",,ott

n, "v Rr - _:- vtotat'LrLrIette.ri'e

Syllabus Objective:

a, recall and use appropriate symbols as sef ouf in Sl Units, Sigr,s, Symbors and

Abbreviations (ASE, 1981) and Signs, Symbols and Systematics (ASE, 1995).

4 Draw the symbols for the following components.

Symbol Description Symbol Description

Fixed resistor ? Electric bell

Variable resistor Heater

Voltmeter Thermistor

Ammeter Potential divider

Galvanometer Loudspeaker

-lt Cell Lamp

l'- - -1, Battery {F Capacitor *

Power supply--1r-.{ll-__-)tL

Transformer with

core

Diode l\,4icrophone *

--€F Fuse Motor *

Y Aerial *Light emitting

diode

(LED)

m) Light Dependent

Resistor (LDR)I Earth connection

b. draw and interpret circuit diagrams containing sources, swifches, resistors, ammeters,

voltneters, and/or any other type of component referred to in the syllabus.

A student has available some resistors, each of resistance 100 O. Draw circuit diagrams, one ineach case to show how a number of these resistors may be connected to produce a combinedresistance of

(c) 40 o(a) 200 o

_-G }-

R=100+'100

=200C)

p=1 L1 l1','r00 r00

=50O

p=1!* L* L1,'r00 r00 )00

=40O

(b) 50 o

(b)(a)

c,

d.

solve problems using the fomula for the combined resistance of two or more resistors in

5e/es,

solve problems using the formula for lhe combined resistance of two or more resistors in

Parcllel.

solve ptoblems involving series and parallel circuils for one source of e.m'f

Three resistors are connected as shown in Figure 6. The points X and Y are connected to a source ofdirect current.

Fis.6

Express the ratio /,4 rn terms ot R- and n, onry. 1| -^^Rt'"lt&

4 = i, + L (Kirchofs First Law)

As R2 and R3 are connected in parallel,V, =V"trR, - 1,R,

. t"R,.R,

1. =!&,1"'R?

l_ &+&l. R,

7 In the circuit shown in Figure 7, a potential difference of 3 V is applied across XY.

6() 2A

Fig.7

Calculate the current passing through the 5 O resistor. [0.375 A]

Effective resistance across XY = 9=4O2

Total current flowina ttrrouqn xY =9 - 0.25 n4

Current llowrng through the b O resrstor = !p.rS 4 -O.SZS e2

ln the circuit shown in Figure B, each of the resistors X and Y has resistance 6 O. The cell C hase.m.f. of 12 V and internal resistance is 3 O.

YFig.8

What is the current passing through Y? [1 A]

Take note that the internal resistance of the cell has to be considered and that the intemalresistance of the cell is not considered to be parallel to X and Y because it does not have the samepotential across it when compared with X and Y.

1EffecLive resistance ot X and Y= - - -3Ott

+66Effective resisiance ofthe entire circuit = 3+3 = 6 O

v 't2Ctrrrent l, = = =24R6By Kirchoff's First Law,

l1= lr+ 13

As X and Y have the same potential difference, since they are connected in parallel,

The qiven circuit in Fiqure I is made up of six resistors each of resistance R.

x

Fig.9

What is the equivalent resistance across x and Y? [Fy3]

parallel parallel

0x

Parallel Take note that theeffective resistance of 2parallel resistors of thesame resistance R is RY2.

This becomes Ri/2 parallelwithEffective resistance = 1/3 R

R.

10 What is the equivalent resistance of the neh,vork between a and b in the Figure 10?

I10 Ol x

Fig. 10

Note: The potential at X and Y are the same, therefore we can ignore the middle resistor.

n = t1*1t'= to o'20 20

00

11

(a) 1, J,, , "6c)

LtFis. 11

(b) %d = 1 x (6+5+1)= 12 V

t6d - tzi J-44

/"d=4+ 1 = 5 A (Kirchoff s First Law)

Vcd=5x4=20VVbc= 12 + 20 = 32V

lac= 16 + 5 = 21 A (Kirchoff's First Law)

ve= 21 x1=21V

(c) The 4 O and 6 O resistorc parallel to the wire will be bypassed

R.h= 1 + (1/2+ 1/3+ 116)1

=24

1T' \11 l-rlLr

L

l2-a5

+1ll s;|-ou 2l

Consider the combination of resisiors in the Figure 11.

(a) Calculate the resistance bet\reen points a & b. [: O])t

(b) lf the current in the 5 O resistor is '1 A, calculate the potential difference between points a &b. t53 vl

(c) lf the points c, d & e were connected together with a conducting wire of negligibleresistance, calculate is the equivalent resistance beh^/een points a & b. [2 O]

1_v\ln 't Nd-- -)

12

show an understanding of the use of a potential divider circuit as a source ofvariable p'd.

A constant 60 V d.c. supply is connected across two resistors of resistance 400 kO and 200 kO as

shown in Figure 12. What is the reading of the voltmeter given that the voltmeter is not ideal andhas a resistance of 200 kO? [12 Vl

t- -* looro--* 2oo ko.:._..._ |

Let R2 be the effective resistance of ihe voltmeter and resistor in parallel connectionLet V2 be the voltmeter reading

p"=1-L111'=196p9' 200 200 '

x60='12V- 100+ 400

Fig. 12

13 Apotentialdividerisusedtogiveoutputsof2Vand3Vfroma5Vsource,asshowninFigurel3.

+3V

+2V

Fig. '13

Which combination of resistances, R7, R, and R3 gives the correct voltages? Show workings.

R1/ka Rr/kQ R3l/koA1 1 2

82 1 2

c3 2 2

D3 2 3

OV

So,

+3V

+2V

&+&+&Only choice B fits the relation.

14 Figure'14 shows a circuit containing a 30V battery and 6 resistors. The potential differencesacross A, C and D are 22V,8 V and 12 V respectively. Calculate the potential difference acrosseach of the components B, E and F and also the potential at the points U, W, X, Y and Z.

[8 V, 10 V, 10 V, 0 V, 12 V, 20 V, -2 V, -10 \',]

Sketch a qraph to show how the potential varies along the line XZ. Label the graph withappropdate values.

Fig- 14

vYz=30-22=BVVuz=30-B-12=10V

Vw=0+12=12VVx=12+B=20VVy=20-22=-2VV,=-2-8=-1OV

i.e p.d. across B is 8 Vi.e. p.d. across E and F are 10 V.(since it is connected to earth)

Distance-2,10

30v

Distance

15

explain the use of thetmistors and light-dependent resistors in potential dividers to provide

a potential difference which is dependent on temperature and illumination respectively.

The light dependent resistor (LDR) and a 500 o resistor form a potential divider between voltagelines held at + 30 V and - 30 V as shown in Figure 15. The resistance of the LDR is 1000 O in thedark but then drops to 100 o in bright light. What is the corresponding change in potential at x?I30 vl

+30v

-30v

V=+30-(-30)= 60 V Fig. 15

In the dark, the resistance of LDR is '1000 o.

P.d. across 500 o resistor = fraction of60 V

= 5oo ,oo500 + 1000

=20v

The potential ofX=(-30)+20 =-10V

ln bright light, the resistance of LDR is '100 o.

P.d. across the 500 o resistor = fraction of 60 V

= 5!o "66500 +100

=50v

The potential of X=( 30)+50 =+20V

The change in the potential ofX = + 20 - ( - 10)= 30 V

't6 A student decided to build a temperature probe and set up a circuit as shown in Figure 16a- Thebattery has e.m.f. 9 V and negligible internal resistance-

The voltmeter has infinite resistance. The calibration curve for the thermistor is shown in the Figure'16b b 3-:17J

Figure 16b

(a) Suggest why it is necessary to include a fixed resistor in the circuit.

A fixed resistor is included in the circuit to form a potential divider circuit such that the potentialdifference across the thermistor is a fraclion of the battery e.m.f.. Any changes in the thermistorresistance associated with changes in temperature will be indicated by changes in the potentialdifference across the thermistor and hence changes in the voltmeter reading.Without the fixed resistor, the potential difference across the thermistor will always be the same,equal to the p.d. across the cell, I V, i.e. ihe reading on the voltmeter will be constant.

1.11r *-a Jo'c1'L

0.lvc

l'58LFigure 16a

(b) The probe is to be used to measure temperature in the range 0 "C and 30 "c. Use the graph in

Figure 16b to find the resistance of the thermistor when the probe is at 30 "C. Hence calculate thereading on the voltmeter for the temperature of 30 'C. [1.25 kO, 7.2 V]

Resistance of the thermistor = 1.25 kO50

Readinq on voltmeter - xg.O=7.2V- 5 0+ I.-25

(c) When the iemperature of the thermistor is 2.5 "C, the voltmeter reads 5.3 V. The voltmeter has arange 0 to 10 V. Suggest one disadvantage of using this voltmeter in the circuit of Figure 16a fortemperature measurement.

For a temperature range of 2.5oC to 30'C, potential difference across the sko resistor will rangeapproximately from around 5.30 V lo 7.2V. Thus only a small part of the voltmeter range will beused and it is difficult to achieve high precision.

h.

17

recall and solve problems using the principle of the potentiometer as a means of comparing

pote nt ia I d ifferen ces.

A potentiometer has a wire XY of length L and resistance R. lt is powered by a battery of e-m.f. Eand internal resistance r in series with a resistor of resistance 2R. With another cell in the branchcircuit, the null point is found to be U3 from X, as shown in Figure 17.

E,r

cell Fig.17

Express the e.m.f. of the cell in terms of E r, R and L (whichever is necessary).

E,r2R

At null point, thep.d. across thewire here isequalto thee.m.f. of this cell.

Let the e.m.f. of this cell be E'The p.d. across the wire (U3) is a fraction of E (by potential divider principle).

SoRi3

R+2R+rER

3(3R +r)

'18 ln a particular potentiometer circuit shown in Figure 18, the balance length, /was found to be toosmall. / can be increased by

Driver Cell with negligible internalresistance

Fig. 18

changing the driver cell to one with larger e.m.f.connecting a protective resistor in series wjth the galvanometer.adding a resistor in series with the driver cell.replacing the slide wire PQ with one that has higher resistance.

Explain your reasoning.

The answer is C.The balance length is such that the p.d. across the length of wire / is equal to the e.m.f. of the cell

Hence,

E" =V,

Hence, to increase the value of /, while Ev and /po is fixed, the only way is to reduce ypo. This can

be done by adding a resistor in series with the driver cell as the e.m.f. of the driver cell will now be

divided bet\Meen the resistor and the wire, instead ofjust the wire only.

Option A will be decreasing I Option B and D will have no effect on the balance length.

One of the reasons for doing so is to increase the sensitivity of the device, so as to determine a

more precise value for the e.m.f. of Y.

BcD

19 ln Fiqure 19, AB is a 10 Q slide wire, 50 cm long. Er is a 2 V accumulator of negligible resistance Rl

and R2 ?re resistances of 15 O and 5 O respectively. When the keys K1 and K2 are both open, thegalvanometer shows no deflection when AJ is 3'1.25 cm. When keys K1 and K2 are both closed, thebalance length AJ is 5 cm. Calculate

(a) The e.m.f. of cell E, [0.5 q(b) The internal resistance ofthe cell E, I7.5 Ql

(c) The balance length AL wnen n, is op&r/and Kr is close. ['12.5 cm]'\/(d) The balance length AJ when K is cloUe and K is open. [12.5 cm]

50to

When K, is ooened. Y." = " (21 = 0.8 V' - 10+15S;nce K, is opened, no current flows in the lower circuit i.e. e.m.f. E2 is measured.

tt )5F, = Y^ i = :-:-- t0 8t = 0.5 V

50

When Kr is closed, V,qs = 2 V

Since K, is closed, current flows in the lower circuit - Vat= Ez- Ir

y^,= 2121 = o.2y'- 50

(a)

(b)

(c)

Since/R.0.2V e /(5)=0.2 = I=0.215=O.O4A

Er-h=0.2 - 0.5-0.04r=0.2 -.t r=7.54

When lG opened, VnL= Ez= O.5Y

When Kr is closed, YAs = 2 V1 0.5 -.t l=12.5cm502

When K, closed, VeL = 0.2 VWhen K1 is opened, Vre = 0.8 V

L=92 .- t = 12.s cm50 0.8

(d)

20 ln using a simple slide-wire potentiometer circuit, a large protective resistance is sometimesconnected in series with the galvanometer. Why is this done?Explain how (if at all) the presence of this resistance affects(a) the position of the balance point,(b) the precision with which it may be found.

(a) The resistance will have no effect on the position of the balance point. At balance, there is

no current flowing through the galvanometer. Hence, the resistance will not cause achanqe in the potential difference across any hto points and therefore, will not affect the

(b)balance length. Adr,.@'4The precision at which the balance length may be found willSflRase. Before balance, acurrent will flow through the galvanometer. With thethe current flowing through the galvanometer will be

existence gf a protective resistance,reduced.lisa.ffidt

'(l^'s yz.o^nn i ,y'La*',++e-, W *{-qn*g ",r-'*^ 4'^- ,* ?*'"%

Practical Application:

21 A tow ol 25 tlecorative lights, connected in sedes, is connected to a mains tlansformer. When thesupply is switched on, the lights do not work. The owner uses a voltmeter to test the circuit. Whenthe voltmeter is connected across the third bulb in the row, a reading of zero is obtained. Which ofthe following cannot be the only fault in the circuit?

A The filament of one of the other bulbs has broken.B The filament of the third bulb has broken.C The fuse in the main transformer has blown.D There is a break in the wire from the supply to the transformer.

' Explain your reasoninq.

The answer cannot be A, C and D because all these options will give you a zero reading for tbe" third bulb. The answer has to be B. lf B is the only fault, then, the voltmeter will give you a non-' zero reading instead.

Electrical devices are often rated with a voltage and a current - for example, 120 volts, 5 amperes.Batteries, however, are only rated with a voltage - for example, 1.5 volts. Why?

An electrical appliance has a given resistance. Thus, when attached to a power source with aknown potential difference, a definite current will be drawn- The device can be labelled with boththe voltage and the current- Batteries, however, can be applied to a number of devices. Eachdevice will have a different resistance, so the current from the battery will vary with the device. Asa result, only the voltage of the battery will be specified.

Why is it possible for a bird to sit on a high-voltage wire without being electrocuted?

The bird is resting on a wire of a fixed potential. ln order to be electrocuted, a potential difference isrequired. There is no potential difference between the bird's feet.

24Data AnalysisCurrent for a car's electrical system is supplied by the battery when the engine is not running, andby the generator. All the current is supplied at the voltage of the battery usually 12 V or thegenerator (approximately 15.5 V), except the current at the spark-plugs, which is boosted by the

ignition system to as much as 3x104 V as required. A car's electrical system is divided intocircuits, all with different basic functions and different controls. They are: the ignition circuit, thestarter circuii, the charging circuit, the lighting circuit and accessories circuits.

Below is a diagram showing a simplified lighting circuits, a table showing the various lamps theirvoltages and power ratings when working normally, and the other the V-l characteristics of one ofthe lamps.

Revelsilg lights

Specjfications for each type of car light

Side lights

Stop light 12V 1BW

Tail light 12V 5W

Side light 12V

Head light 12V 60w

Reversing light 12V 21W

/

Lamp characteristics

15

p.d.A/

'10

0.5 1.0 1.5 currenvA

(a) State and explain the type of car light the above graph represents.

Stop light. IllThe current at p.d. 12 V is 1.5 A. The power consumption is12v \ 1.5 A= 18 W. (stop light) t1l

(b) Complete the table below.

t3l

Current/A o.2 o.4 0.B 't.0 1.2 1.4 1.6

p.dA/ 0.3 '1 0 2.2 3.6 5.4 10.4 14.O

ResistanceIQ

1.5 2.5 4.5 5.4 6.5 7.4 8.8

(c) Plot a graph of resistance against current for the lamp.(Assume linear relationship)

t3l

R/O9

8

'7

6

5

1

3

2

I

0.2 0.4 0.6 0.8 l.o 1.2 t.4 1.6 I/A

From the graph, obtain the resistance of the lamp when

(i) normally,

8.3 - 8.4 f)

(iD cold. Explain your method for(ii).

0.4 e)When cold, no current flows through the lamp.At I = 0 A, obtain the intercept on the R axis.

tll

t1lt1ll1l

.9

(:

I

&'n*/

6/

(d) Why are all the car lights arranged in parallel?

Two basic reasons:1. All will receive the same p.d. of 12 V lll2. When 1 lamp breaks, other lamps are not affected. I1l

(e) Calculate the toial current drawn from the battery when the car is stationary and the headlights, thereversing lights and the stop lights are on.

2..60Current drawn by headlights = -,;- -10A

current drawn by reversrng lrghts = 3-Z - 3.5 A12

,^60Current drawn by stop lights = - - :- - 10 A

12Total current drawn = 23.5 A 121

(f) Why is ihe initial current through each type of light greater than when it is operating nomally?

lnitially, the lamp is cold when just switched on.Resistance is low. Larger current flows. I1lAt normal operating condition, the lamp is hot andhigher resistance. Lower current flows consequently. 11]

(g) The function of the spark plug is to provide a spark so that the air-petrol mixture can be ignited.Suggest how the spark could be created in the ignition system.

High potential ofthe spark plug ionizes the air in the cylinder. I1lElectrons and air ions flow resulting in sparks created. nl

(h) Suggest one other circuit, other than the lighting circuit, that may draw current from the battery.

Accessoies circuit (e.q. radio, CD player), ignition circuit, etc. 11l

*** End ofTutorial ***