Category: Physic Tutorials

Notes on electric field for Cambridge A level and UTME

Electric field is a region of space where an electric charge body/object experience a electric force.

Field of force

gravitational field – point mass

electric field – charge

magnetic field – current carrying conductor / charge

Field of force are always vector quantities because their field lines always show the direction. Electric field lines will always go in the direction from the positive to the negative region.

Positive point charge : the field lines are radially outward and are always in the direction of the force

Negative point charge: the field lines are radially inward

Point charge are spherical in shape and infinitesimal in nature

Electric field strenght: at a point in an electric field field is defined as the force per unit charge acting on a positive stationary charge at that point

E = f / q

cross multiply

F = qE

A ‘test charge’ should be positive. the direction it moves in then shows the direction of the electric field.

Uniform electric field

1.The field lines must be parallel

2.The field lines must be equally spaced

The electric field lines is from region of higher potential to a lower potential

The force acting on a charge inside a uniform electric field is constant at all point

The path of a charge inside a uniform electric field is always parabolic in path

Example of a uniform electric field is when you have two parallel plate of

E = v / d

E is the electric field strenght

v is the electric potential difference

d is the distance between the plates

Note that E can also be f / q

E = v / d = f / q

cross multiply

f.d = v . q

f . d = workdone

workdone = qv

Moving charge inside a uniform electric field

But a positive charge will deflect toward the negative charge in a parabolic path

Ek = qv

Ek = 1/2mv2

equate the two

1/2 m v2 = qv

Vel is the speed of the charge

v is the potential difference

m is the mass of the charge

q is the charge

Recommended: Solved questions on electric field

 

How to solve questions on Nuclear Physics for Cambridge A level

Question 1

The gold nucleus 185Au79 undergoes alpha decay. What are the nucleon number and proton number of the nucleus formed by this decay?{cambridge A level may/june 2016 p12}

Solution

Au undergoes alpha decay

185Au79 – 181X77 + 4He2

The above equation is conserved since the number of nucleon and proton are the same before and after decay

The nucleon numberof the nucleus formed is 181 and the proton number is 77

C is the correct option

Question 2

In a time of 42.0 minutes, the count rate from the sample of copper-66 is found to decrease from 3.62 × 104 Bq to 1.21 × 102 Bq. Calculate the half-life of copper-66.{cambridge A level may/june 2016 p41}

Solution

Question 3

The diagram shows part of a radioactive decay chain in which the nuclide thorium-232 decays by α-emission into radium-228. This nuclide is also unstable and decays by β-emission into a nuclide of actinium. This process continues.{cambridge A leve oct/nov 2015 p11}

What are X, Y and Z?

Solution

since Ra undergoes beta decay X will still be 228: beta decay does not affect nucleon number

Y is beta since Th nucleon number isnt affect and the proton increase by 1

Z is Ra since the proton number is 88, after Th has undergone alpha decay, having the same proton number with Ra making it an isotope of Ra

B is the correct option

Question 4

In a model of a copper atom of the isotope 63Cu29, the atom and its nucleus are assumed to be spherical. The diameter of the nucleus is 2.8 × 10–14 m. The diameter of the atom is 2.3 × 10–10 m.{cambridge A leve oct/nov 2015 p22}
Calculate the ratio

Solution

denisty of nucleus = mass of nucleus / volume of nucleus

volume of nucleus =

Volume = 4/3*3.142*(2.8×10^-14)^3

mass of nucleus = 63u

density of atom = mass of atom / volume of atom

volume = 4/3*3.142*(2.3×10^-10)^3

mass of atom = 63u

mass of atom and mass of nucleus (approx.) equal

ratio

ratio = 5.5 x 10 ^11

Question 5

In the D-T reaction, a deuterium (2H1) nucleus fuses with a tritium (3H1) nucleus to form ahelium-4 (4He2) nucleus. The nuclear equation for the reaction iscambridge A level 2014 p41}

Some data for this reaction are given in Figure below

Use data from Figure above to determine the energy released in this D-T reaction.

Solution

total mass of the reactant = 2.01356 + 3.01551 = 5.02907 u

total mass of the product = 4.00151 + 1.00867 = 5.01018 u

mass defect = total mass of reactant – total mass of product = 5.02907 u – 5.01018 u

mass defect = 0.01889 u

note

1u is equivalent to 934 Mev

so,

0.01889 u = (933 * 0.01889) Mev

energy released in this D-T reaction = (934 * 0.01889) Mev = 17.6 Mev

Recommended: Question and answer on Nuclear for UTME

Recommended: Short notes on Nuclear physics

Notes on nuclear physics for Cambridge A level and UTME

Rutherford alpha particle scattering experiment: Experimental evidence for nuclear atom:Results of an experiment where a beam of alpha particles is fired at a thin gold foil (about 1µm thick): where n= number of alpha particles incident per unit time.

Alpha particle are helium atom particles. He bombarded positive alpha particle on thin foil of gold approximately 8.6 x 10^-6 cm thick and took observations of the screen Zns which was behind the gold foil.

A gold foil was used because gold can be made into a very thin sheet or foil

Reasons why beta particle from a radioactive source would be inappropriate for this type of scateering experiment

1. beta particles have a range of energies
2. beta particles deviated by orbital electron
3. beta particle has very small size

Note

Change in Isotope of gold doesn’t affect deviation because deviation depends on charge on the nucleus or electrostatic repulsion i.e same charge since isotopes are element of the same proton number but different nucleon number, so no change in deviation

Observations

1. Most of the α-particles passed through the metal foil undeflected or deflected by(deviated through) very small angles less than 90
2.  A very small proportion was deflected by large angles more than 90°( some of these approaching 180°)

Conclusion

1. the nucleus occupies only a small proportion of the available space in comparism with atom size (i.e the atom is mostly empty space)
2. the nucleus is very small and heavy/dense/massive and +vely charged (since the positively-charged alpha particles are repelled/deflected).

 The energy conversion in the α-particle scattering experiment:

The kinetic energy of the incoming α-particle is converted to the electrical potential energy when it stops at the point of closest approach and turns around.

So  initial kinetic energy of α + nucleus(= 0) =  maximum electrical potential energy of both particles                                                                                           i.e

1/2m_α v_α² =  Q_α .Q_{N /}4πεr

where r is the distance( of closest approach) of α-particle to the nucleus,

Q_α = 2e,       and      Q_{N =} Ze  ,  Z is proton number

Reasonable estimates:

*Nuclear diameter ≈10^-15m …10^-13  to  10^-15 m                                                                                                                        *Atomic diameter ≈ 10^-10-m   …..10^-9  to  10^-11m

Nucleon:   A particle within the nucleus; can be either a proton or a neutron;they are subatomic particles.

Nuclide:   An atom with a particular number of protons and a particular number of neutrons

Proton number Z {old name: atomic number}: Number of protons in an atom

Nucleon number N {mass number}: Sum of number of protons and neutrons in an atom

Isotopes: are nuclei/atoms with the same proton number, but different nucleon/neutrons number

Density Calculation

Density = mass / volume

mass is mass of nucleus

v is the volume of nucleus

r is the radius of the nucleus

mass of a nucleus is measure in atomic mass unit because it is very small

Nuclear Energy

               Mass Defect                                                                                                                                          

Whenever a reaction results in a release of energy, there is an associated decrease in mass, called mass defect, which is converted to energy(in form of gamma radiation with c = 3.00 × 10⁸m/s).

The mass of a nucleus is always less than the total mass of its constituents (protons plus neutrons).

Mass  defect is this difference between the mass of a nucleus and the total mass of its individual nucleons,

i.e

 

Mass defect = (total)mass of nucleons ― (single)mass of nucleus

= Zm_p + (A – Z)m_n – Mass ofNucleus

Mass defect =  final mass – initial mass

Nuclear Binding Energy:

• This is energy that is required to completely separate the nucleons in a nucleus.

OR

The energy released (not  energy lost) when a nucleus is formed from its constituent nucleons

The Binding Energy per nucleon is a measure of the stability of the nucleus since it represents the average energy needed to remove a nucleon. The higher the binding energy, the more stable the nucleus and vice versa

Energy & Mass are Equivalent and inter-convertible.

Thus, Binding Energy, the energy released during nuclear reaction is

         E(J) = Mass defect(kg) × c²

Mass defect of a nucleus is the difference between the toatal mass of a separate necleus and the combined mass of the nucleus

         E = Increase/decrease in binding energy

                i.e  Energy released = total B.E after ― total B.E before

So

I u = 931Mev

         E(Mev) = 931Mev/u × Mass defect(u)   

         E = Binding energy per nucleon × nucleon number

if binding energy is:                                                                                                                                                       

+ve,nucleus is stable, energy released appears as k.e of products    

                                                                                                                          

  ―ve, nucleus is unstable and will decay spontaneously, energy is needed to produce reaction    

Nuclear fission:

The disintegration of a heavy nucleus into two lighter nuclei of approximately same mass. Typically, the fission fragments have approximately the same mass and neutrons are emitted

Nuclear fusion:

The joining together of two light nuclei of nearly equal mass to form a heavy nucleus.

Radioactive decay:

Radioactivity is the spontaneous and random decay of an unstable nucleus, with the emission of an alpha or beta particle, with or without  the emission of a gamma ray photon.

Spontaneity:

*The emission is  unaffected (not speeded up or slowed down) by factors outside the nucleus,  e.g

1. chemical reactions(acids),
2. environmental factor/surrounding (heat, wind, cold, earth, pollution, weather)
3. all external factors.

Randomness:

*It cannot be predicted when the next emission will occur/which particular nuclei will decay next

*All nuclei have equal chance of decay i.e constant probability of decay per time of a nucleus

       

 

Exponential Decay curve

For a large number of a particular nuclei/species, the rate of decay is directly proportional to the number of parent nuclei present.

i.e                                         ― dN/dt  α  λ

dN/dt  =  ―λN                                         (Activity defined in Bq)

λ = +ve constant of proportionality/radioactive decay/disintegration constant in s^-1

 

where N= number of undecayed/active/parent nuclei at that instant/remaining.

 i.e the number of nuclei N remaining and/or activity after some time  t,  decreases exponentially.                                                                                          

              Units:     1Bq = 1disintegration per second,  1Ci = 3.7 ×10¹⁰s^-1

 

*Decay constant λ is defined as the probability of decay of a nucleus per (unit) time

*Activity is defined as the rate at which the nuclei are disintegrating/decay rate i.e number of decays per time (―ve…decreasing)

A = dN/dt = ― λN,      and          A₀ = λ N₀

Since number of undecayed nuclei ∝ Mass of sample,

Number of nuclei in sample = (Sample Mass / Mass of 1 mol) x N_A

Also, following the decay law,

N = N₀e^–λt ,             A = A₀e^–λt ,              C = C₀e^–λt

                                                                                                                                     

 

Half-life ,T_1/2      

  Half-life is defined as the (average) time taken for half the number {not: mass or amount} of undecayed nuclei in the sample to disintegrate,

or, the (average) time taken for the activity to fall to half of its original value.

T_½ = (ln2) / λ

                                                             or              T_1/2 = 0.6931/λ

Recommended: Question and answer on Nuclear for UTME

This effect was first noted by Christian Doppler in 1842. The effect is widely used to measure velocities, usually by the reflection of a transmitted wave from the moving object, ultrasound for blood in arteries, radar for speeding cars, and thunderstorms. The velocities of distant galaxies are measured using the Doppler effect (the redshift).

What is Doppler effect? – It is the apparent change in frequency when there is a relative motion between the source and the observer.

As the source approaches

As the source moves away or recedes

V is the speed of sound

Vs is the speed of the source

Fs is the frequency of the source

F’ is the observed frequency by the stationary observer

Related Article: Note on direct current

The Doppler effect for electromagnetic waves such as light is greatly used in astronomy and results in either a redshift or blueshift. It has been used to measure the speed at which stars and galaxies are approaching or receding from us; that is, their radial velocities.

Red and blue shift: As the star moves away from the Earth, there will be an increase in wavelength and a decrease in frequency; this is redshift. As the star moves away towards the earth, the wavelength will decrease, and frequency will increase; this is blueshift.

Notes on Direct Current for Cambridge A level and UTME

Direct current : flow of charges in the circuit is in the same direction all the time from higher potential to lower potential.

Appropriate circuit symbol

1. switch

2. Cell, voltmeter, ammeter, resistor, variable resistor etc

Circuit diagram

Kirchhoff’s First Law: The sum of the currents entering a junction is always equal to the sum of the currents leaving it(conservation of electric charge)

Kirchhoff’s Second Law: The sum of the electromotive forces in a closed circuit is equal to the sum of the potential differences (conservation of energy)

Derive, using Kirchhoff’s laws, a formula for the combined resistance of two or more resistors in series

Series arrangement

Resistors are said to be connected in series if current can flow from one resistor to another without branching.

V = V1 + V2 + V3

V1 = IR1

V2 = IR2

V3 = IR3

V = IR

IRt = IR1 + IR2 + IR3

Note

in series arrangement same current flows through the resistors but different voltage

Rt = R1 + R2 + R3

Parallel arrangement

When resistors are connect in parallel current branches out but the voltage are the same for all the resistors

I = I1 + I2 + I3

I1 = V / R1

I2 = V / R2

I3 = V / R3

I = V / Rt

V / Rt = V/R1 + V/R2 + V/ R3

1 / Rt = 1 / R1 + 1 / R2 + 1 / R3

Example

Two cells of e.m.f. E1 and E2 and negligible internal resistance are connected into the network in the figure below

The currents in the network are as indicated in Fig. above
Use Kirchhoff’s laws to state the relation
(i) between currents I1, I2 and I3,

(ii) between E2, R, I2 and I3 in loop BCXYB,

(iii) between E1, E2, R, I1 and I2 in loop ABCXYZA.

Solution

According to Kirchhoff’s First Law: The sum of the currents entering a junction is always equal to the sum of the currents leaving it

Let junction B be our reference point

I1 is entering

I3 is leaving

I2 is entering

I3 = I1 + I2

(ii)

loop BCXYB

following anti-clockwise loop

conventionally current moves from positive terminal and “enters” negative terminal

But in E2 current leaves the negative terminal i.e E2 will be negative

I2 and I3 will also be negative because it opposes the direction of the loop

-E2 = -I2R – I3R

E2 = I2R +I3R

(iii)

loop ABCXYZA

following the anti-clockwise loop

Current leaves negative terminal of E2( this will give us -E2) and enters the negative terminal of E1( this will give us +E1)

I1 will be positive because it is in the direction of the loop

Resistors on A and Z are in series so current I1 flows without branching

I2 will be negative because it opposes the direction of the loop

E1 – E2 = I1R + I1R – I2R

E1 – E2 = 2I1R – I2R

Show an understanding of the use of a potential divider circuit as a source of variable p.d.

Vin = V1 + V2

V1 = IR1

V2 = IR2

Vin =IR1 + IR2

Vin = I(R1 +R2)

I = Vin / (R1 + R2)

V2 = Vin / (R1 + R2) * R2

V2 = Vout

Vout = R2 / (R1 + R2) * Vout

Use of thermistors and light-dependent resistors in potential dividers to provide a potential difference that is dependent on temperature and illumination respectively

Thermistor as temperature sensor

Temperature against the resistance of a thermistor

 

Recommended: Questions and answers on direct current for A level and UTME

How to solve questions on stationary waves, diffraction and interference

Question 1

A parallel beam of light of wavelength 450 nm is incident normally on a diffraction grating which has 300 lines / mm. What is the total number of intensity maxima observed?{ Cambridge A level may/jun 2016 p11)
A 7                             B 8                   C14 D15

Solution

d is the slit separation in meters

300 line  is 1 mm

1 mm is 10^-3 m

1 line = 10^-3 / 300 m

d = 10^-3 / 300

wavelength = 450nm = 450 x 10^-9 m

the angle of diffraction = 90 in order to calculate the total number of maxima

n x 450 x 10^-9 = 10^-3 / 300 x sin 90

n = 7, which is the number of order

the total number of maxima = 2 x 7( 7 order up and 7 order down) + 1(zero order) = 15

D is the correct option

 

Question 2

Wave generators at points X and Y produce water waves of the same wavelength. At point Z, the waves from X have the same amplitude as the waves from Y. Distances XZ and YZ are as shown.

When the wave generators operate in phase, the amplitude of oscillation at Z is zero. What could be the wavelength of the waves?{ Cambridge A level May/jun 2015 p13}
A 2 cm                        B 3 cm       C4cm                           D 6 cm

Solution

since the amplitude at z is zero, it means it is a destructive interference

yz – xz = (n + 1/2)ℷ

34 -24 = (n + 1/2)ℷ

10 = (n + 1/2)ℷ

n =1, 2, 3,4, ……..

using n = 2

10 = 2.5 ℷ

ℷ = 10/2.5

ℷ = 4 cm

Question 3

A parallel beam of white light passes through a diffraction grating. Orange light of wavelength 600 nm in the fourth order diffraction maximum coincides with blue light in the fifth order diffraction maximum. What is the wavelength of the blue light? {Cambridge A level may/jun 2014 p11}
A 450 nm              B 480 nm                       C 500 nm                           D 750 nm

Solution

when one light coincides with another in diffraction grating

600 * 4 = 5 * ℷ

ℷ = 480 nm

B is the correct option

Question 4

The basic principle of note production in a horn is to set up a stationary wave in an air column.

For any note produced by the horn, a node is formed at the mouthpiece and an antinode is formed at the bell. The frequency of the lowest note is 75 Hz. What are the frequencies of the next two higher notes for this air column?

{Cambridge A level may/jun 2014 p11}

Solution

a node is formed at the mouthpiece and an antinode is formed at the bell means, one end is closed and the other end is opened.

for the lowest frequency(first harmonic)

ℷ =4l

v = fℷ

v = 4lf

f = v / 4l

fo = 75 Hz

for second harmonic

L = 3ℷ / 4

ℷ = 4l / 3

v = fℷ

v = 4lf / 3

f = 3v / 4l

f = 3f0

f = 3 * 75 = 225 Hz

For 3rd harmonic

f = 5fo

f = 5* 75 = 375 Hz

D is the correct option

Question 5

Fig below shows a string stretched between two fixed points P and Q.

A vibrator is attached near end P of the string. End Q is fixed to a wall. The vibrator has a frequency of 50 Hz and causes a transverse wave to travel along the string at a speed of 40 m s–1.

Calculate the wavelength of the transverse wave on the string.{ Cambridge A level may/june 2013 p22)

Solution

v = f ℷ

40 = 50 * ℷ

ℷ = 40 / 50

ℷ = 0.8 m

Question 6

Light of wavelength 600 nm is incident on a pair of slits. Fringes with a spacing of 4.0 mm are formed on a screen. What will be the fringe spacing when the wavelength of the light is changed to 400 nm and the separation of the slits is doubled? { Cambridge A level may/ june 2013 p 11}
A 1.3 mm                  B 3.0 mm                                   C 5.3 mm                                 D 12 mm

Solution

D is constant

600 / 4*1 =  400 / 2x

x = 4 * 400 / 600 * 2

x = 1600 / 1200

x = 1.3 mm

Recommended: Note on stationary waves, diffraction grating and interference

In this article, I will be discussing stationary waves, diffraction and interference

Principle of superposition

The Principle of Superposition states that when two or more waves meet at a point, the resultant displacement at that point is equal to the sum of the displacements of the individual waves at that point.

For constructive interfernce: the resultant displacement is always higher than the displacement of the individual waves.

For destructive interfernce:the resultant displacement is always smaller than the displacement of the individual waves.

Stationary waves

A stationary wave is formed by two progressive waves of the same type, amplitude and frequency travelling in opposite directions superpose.

show an understanding of experiments that demonstrate stationary waves using microwaves, stretched strings and air columns

Microwaves

The Microwave source generate the wave signal while the metal reflector reflect the wave back, by this there the two progressive waves will superpose.

Stretched string

Explain the formation of a stationary waves using a graphical method, and identify nodes and antinodes

Read: Overtone in closed and open pipe

Node: it is a point of zero amplitude

Antinode: it is a point of maximum amplitude

Recall,

Differences between stationary and progressive waves

staionary waves

1. No energy is transported by the wave
2. the wave profile does not advance
3. amplitude varies from maximum at the anti-nodes to zero at the node

Progressive waves

1. energy is transported in the direction of the wave
2. wave profile advances
3. amplitude is the same for all particles in the wave

Diffraction

Diffraction is the spreading of waves through an aperture or round an obstacle.

• The bending of waves around an obstruction
• As the size of the aperture or the object decreases the effects of diffraction increase
• The wavelength needs to be similar to the size of the aperture for diffraction to be noticeable

The smaller the size of the aperture, the greater the spreading of the waves (if the width of the aperture is about the same size as the wavelength, λ, the diffraction effect is very considerable).

Diffraction grating

A diffraction grating is a plate on which there is a very large number of identical, parallel, very closely spaced slits.

Uses of diffraction grating

1. it is use to determine the wavelength of light
2. it can be used to make spectrometer

Diffraction grating equation can be represented by

n is the number of order

λ is the wavelength

d is the slit sepation

ø is the angle of diffraction

Interference

Interference is the superposing of two or more waves to give a resultant wave whose displacement is given by the principle of superposition.

At regions of maxima, constructive interference occurs (i.e the waves arrive at these points in phase), resulting in maxima amplitude, hence high intensity

At regions of minima, destructive interference occurs (i.e the waves arrive at these points in anti-phase), resulting in minima or zero amplitude, hence low or zero intensity.

Young’s double slit experiment

Monochromatic light talks about light with one wavelength, example red light

polychromatic light consists more than one wavelength, example visible or white light

Coherent source: waves coming from them are always at a constant phase difference

Bright fringes are formed due to constructive interference(i.e the waves arrive at these point in phase), while

Dark fringes are due to destructive (i.e the waves arrive at these points is anti-phase(180 degree)- no resultant amplitude, which then appear dark)

For interference fringes to be observable

1. The source must be coherent; that is they must maintain a constant phase difference
2. The source must have the same frequency (for light waves, this mean that they must be monochromatic)
3. The pronciple of superposition must apply(the source must produce the same type of waves)
4. The source must have approximately the same amplitude

Condition for constructive interference

B is the second source

A is the first source

Condition for destructive interference

λ is the wavelength

a is the separation of the slits

x is the fringe separation

D is the separation between the screen and the double slit

Recommended: Solved questions on waves

How to solve questions on current of electricity for Cambridge A level

Question 1

A heater is made from a wire of resistance 18.0 Ω and is connected to a power supply of 240 V. The heater is switched on for 2.60 Ms.{Cambridge A level 2013 p22}
Calculate
(i) the power transformed in the heater,

(ii) the current in the heater,

(iii) the charge passing through the heater in this time,

(iv) the number of electrons per second passing a given point in the heater.

Solution

i.   P = v2 / R = 240 *240 / 18 = 3200 watt

ii.   v = IR

I = v / R = 240 / 18 = 13.3 A

iii.  Q = It

Q = 13.3 * 2.6 x 1o^6 = 34.7 x 10^6 c

iv  It = ne

n/t  = I/e  = 13.3 / 1.6 x 10^-19 = 8.33 x 10^6  s-1

Recommended: More solved questions on current of electricity

Question 2

An iron wire has length 8.0 m and diameter 0.50 mm. The wire has resistance R. A second iron wire has length 2.0 m and diameter 1.0 mm. What is the resistance of the second wire?{Cambridge A level 2012 p11}
A. R / 16                   B. R/8                    C. R / 2                    D. R

solution

Since they are of the same material, they will have the same resistivity

R1A1 /L1  = R2A2 / L2

R * 0.5 * 0.5 / 8 = R2 * 1 * 1 / 2

R2 = 2*R*0.5*0.5 / 8

R2 = R / 16

A is the correct option

Question 3

An electric heater is to be made from nichrome wire. Nichrome has a resistivity of 1.0 × 10–6 Ω m at the operating temperature of the heater. The heater is to have a power dissipation of 60 W when the potential difference across its
terminals is 12 V.
(a) For the heater operating at its designed power,
(i) calculate the current,

(ii) show that the resistance of the nichrome wire is 2.4 Ω.

(b) Calculate the length of nichrome wire of diameter 0.80 mm required for the heater.

(c) A second heater, also designed to operate from a 12 V supply, is constructed using the same nichrome wire but using half the length of that calculated in (b). Explain quantitatively the effect of this change in length of wire on the power of the heater. {Cambridge A level 2010 p21}

Solution

P = Iv

60 = I * 12

I = 60 / 12

I = 5 A

ii.

V = IR

R = v / I

R = 12 / 5

R = 2.4 Ω

b.

R =  pl /A

l = RA/ p

l =( 2.4 * 3.142 * 0.8 x 10^-3 * 0.8 x 10 ^-3) / 4 * 1 x 10^-6

L = 1.21 m

C.

The resistance and the length are directly proportional

when the length is halved, the resistance will be halved

resistance and current are inversely proportional

reduction in value of resistance means more current will flow. since the resistance has been halved, the current will be doubled.

since second heater also operated on a 12v supply, the power will be doubled because the current has been doubled i.e they are directly proportional.

Question 4

A source of e.m.f. of 9.0 mV has an internal resistance of 6.0 Ω. It is connected across a galvanometer of resistance 30 Ω. What will be the current in the galvanometer?{Cambridge A level 2010 p11}
A 250 μA              B 300 μA                    C 1.5 mA                           D 2.5 mA

Solution

E.m.f = I(R + r)

9 x 10^-3 = I(6 + 30)

0.009 = 36I

I = 0.009 / 36

I = 0.00025 A = 250μA

A is the correct answer

Question 5

A 12 V battery is charged for 20 minutes by connecting it to a source of electromotive force (e.m.f.). The battery is supplied with 7.2 × 104J of energy in this time. How much charge flows into the battery?{Cambridge A level 2009 p1}
A 5.0 C                   B 60 C                             C 100 C                            D 6000 C

Solution

Energy = power x time

power = Iv

E = Ivt

Q = It

E = Qv

Q = E /v

Q = 7.2 x 10^4 /  12

Q = 6000 C

D is the correct answer

Recommended: Note on current of electricity

Notes on current of Electricity for Cambridge A level and UTME

Current of Electricity

Electric charge Q passing a point is defined as the product of the (steady) current at that point and the time for which the current flows.

Q = It

Q = ne

I is the current

t is the time taken

When charges flow, there is electric current  I, therefore,

Electric current is the rate of flow of electric charge

I = ΔQ/Δt

The conventional direction of current is that of the +ve charge i.e current moves from the positive terminal to the negative terminal

One coulomb: is defined as the charge flowing per second past a point at which the current is one ampere.

Potential difference: Potential difference is defined as the energy transferred from electrical energy to other forms of energy e.g heat, light, sound e.t.c, when unit charge passes through an electrical device.

W = QV

W is the workdone

Q is the electric charge

v is the potential difference

V = W/ Q, the unit is JC-1 or volt

The volt: is defined as the potential difference between two pts in a circuit in which one joule of energy is converted from electrical to non-electrical energy when one coulomb passes from one point to the other, i.e 1 volt = 1 JC-1.

Resistance: is defined as the ratio of the potential difference across a component to the current flowing through it, provided temperature( and other physical conditions like resistivity, area of cross section, and length of wire remains constant.)

The Ohm: is the resistance of a resistor if there is a current of 1 A flowing through it when the p.d across it is 1 V, i.e,
1 Ω = One volt per ampere

R = V/I

Power is the rate of energy expended

P = E / t

E = Pt

V = W/Q

Substitute for E

V = Pt / Q

Q = It

t = Q / I

Substitute for t

V = PQ/IQ

V = P /I

P = IV ……….i

Note

V = IR

substitute for V

P = I^2 R

sketch and explain the I-V characteristics of a metallic conductor at constant temperature, a semiconductor diode and a filament lamp

For Metallic conductor

For filament lamp

For semiconductor

Ohm’s Law:  The current flowing through a piece of metal is proportional to the potential difference across it providing the temperature remains constant

            From the ohm’ law equation for ohmic conductors,

(1) as R increases, the p.d drawn by the load increases, and vice versa, but R is inversely proportional to I, so current decreases because, as resistance increases, current decreases and vice versa.

(2)for a resistor and/(variable) resistor, as the R of one decreases, the p.d of the other increases, its current also increases, because, the

(3)V/I is always a constant value at the same temperature

 

Resistivity

Resistivity is defined as the resistance of a material of unit cross-sectional area and unit length.

R = Resistance

r = Resistivity of material

L = Length of conductor

A = Area

E.m.f and P.d

Electromotive force Ɛ is defined as the(total) energy transferred / converted from non-electrical forms into electrical energy when unit charge is moved round a complete circuit.

distinguish between e.m.f. and p.d. in terms of energy considerations

e.m.f.  = (energy converted from other forms to electrical) / charge

p.d. = (energy converted from electrical to other forms) / charge

Internal resistance

Internal resistance is the resistance to current flow within the power source. It reduces the potential difference (not  the emf) across the terminal of the power supply when it is delivering a current.

Consider the circuit below

E.m.f = I(R +r)

E.m.f = IR +Ir

IR is the terminal P.d

Ir is the lost volt

The greater the internal resistance, then the greater percentage loss in energy per unit charge, and the lower the terminal p.d V( = IR), and vice versa. So reducing the internal resistance of a battery increases the effective/useful energy delivered per unit charge across the external load.

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