Category: Physic Tutorials

Notes on basics electronics for UTME

Conductors are substances that allow the passage of electricity through them. Insulation is a substance that will not permit the passage of electricity through them.

Insulators are material that do not allow the passage of electricity through them.

Semi conductors are materials having intermediate electrical conductivity between that of conductors and insulators.

Examples of Semiconductors

1. Silicon 2.      Germanium

Intrinsic semi conductors are semi conductors in their pure state i.e semiconductors that have not be doped.

Extrinsic semiconductors are semiconductors that have been doped.

Doping is the addition of impurities such as antimony, and germanium crystal to a semi-conductor. The aim of doping is to alter the structure of the semiconductor crystal.

Current is carried through a semiconductor by two types of carriers: (i) free electrons which have negative charges (ii) holes that have a positive charge

In semiconductors such as silicon and germanium,there are four electrons in the outermost shell – valence electrons.

Effects of Temperature on Semiconductors

The resistance of a semiconductor will be decreasing as the temperature increases. Unlike, in the case of a pure conductor where the temperature increases with the rise in conductivity.

Types of semiconductors

1. n-type semiconductor
2. p-type semiconductor

n-type semiconductor: it is the addition of pentavalent element such as antimony, and germanium crystal to a semi-conductor will lead to the formation of the n-type of the semiconductor. The majority carriers are electrons and minority carriers are holes.

p-type semiconductor is produced by “doping” pure germanium or silicon crystal with an impurity or elements that have three valence electrons(trivalent),e.g., boron and indium. the majority carriers in the p-type are the positively charged holes.

The p-n junction diode

The p-n junction semiconductor is a single semiconductor made up of a p-type semiconductor and n-type semiconductors. The two types form an alloy with a fragile junction between the two semiconductor types.

Reverse biase

In reverse biase the p-type is connected with the negative terminal while the n-type is connected to the positive terminal. This make the depletion layer to widen.

Forward biase

In forward biase the p-type is connected to the positive terminal of the battery while the n-type is connected to the negative terminal of the battery. This make the depletion layer to narrows.

Uses of p-n junction diode

1. As a rectifier for alternating current and direct current voltage because its resistance can be varied from the forward biased to the reverse biased.
2. The potential difference requires to operate it is small,e.g., in a radio receiver, one can use 3v of battery.
3. Cheaper to manufacture
4. Smaller in size

Recommended: Solutions to some UTME Physics questions

Notes on alternating Current through a Capacitor and Inductor

A.C through a capacitor

An alternating voltage V = Vo sin wt when applied across a capacitor C, the current is given by:

Both the experiment and theory show that the voltage (V) and the current (I) are out of phase. The current leads the voltage by 90 degree i.e. the current is ahead by the voltage. The voltage is said to lag on the current.

Capacitance Reactance Xc

The opposition offered by the capacitor to the flow of alternating current is known as the capacitance reactance Xc.

Peak and r.m.s values of A.C

The amplitude or peak value of the current Io is the maximum value of the current. The root mean square is that steady current that will develop the same quantity of heat, at the same time, and at the same resistance.

Inductance in Alternating Current Circuit

When an inductor is connected to an alternating voltage source, V_L = V_o sin wt and the current

I is given by I_L = I_osin (wt)

The current is delayed behind the voltage in the circuit i.e the current lags the voltage by 90 degree.

Where the unit of L is Henry (H), f is in Hertz and X_L in ohms

Resistance (R) and, Inductor (L) series circuit

Capacitor- Resistor (C-R) series circuit

Series Circuit containing Resistance (R), Inductance (L) and Capacitance C

Example1

In an A.C circuit, if the peak voltage and current are 330volts and 80A what is the value of the root mean square voltage and current

Solution

Example 2

A series circuit consists of a resistance 600 ohms and an inductance of 5 Henry’s An A.C voltage of 15 volts (r.m.s) and frequency 50Hz is applied across the series circuit calculate

1. The current flowing through the circuit
2. The voltage across the inductor

solution

Recommended: Solved questions on alternating current

 

The surface of the liquid behave like a membrane under tension, surface tension arises because the molecules of the liquid exert attractive forces on each other. There is zero net force in a molecule inside the volume of the liquid, but a surface molecule is drawn into the volume. Thus, the liquid tend to minimise its surface area, just as a stretched membrane does.

Three examples of surface tension

1. Water striders spend their life walking on water
2. Surface tension causes the drops at the ends  of these pipettes to adopt a near-spherical shape
3. A paper clip rests on water, supported by surface tension

Surface Tension Dimensional Formula and Unit

Coefficient of surface tension is defined as the force per unit length acting in the surface at right angle to one sides of a line drawn in the surface. The s.i unit of surface tension is Nm^-1 (Newton per meter).

Dimension

y = F/L

The dimension of force is MLT^-2

The dimension of length is L

Y = MLT^-2/L = MT^-2

So the dimension of coefficient of surface tension is MT^-2.

Adhesion and Cohesion

Cohesion is the force of attraction between molecules of the same kind. Adhesion is the force of attraction between molecules of different kind.

Cohesion and Adhesion explain the different action of water and mercury when spilled on a clean glass surface.

Water: The adhesion of water molecules to glass is stronger than the cohesion between water molecules, water spread out on a clean glass surface when sprinkled on it and wet the glass.

Mercury: The force of cohesion between two molecules of mercury is greater than the force of adhesion between a molecule of mercury and a molecule of glass; thus mercury gathers in pools when split on glass.

Angle of contact

The angle of contact is the angle between the tangent to the liquid surface and the tangent to the liquid surface, both drawn in the place at right angle to both surfaces and meeting at the point of contact measured within the liquid.

For a liquid in which the cohesive force is greater than the adhesive force, the angle of contact is obtuse e.g mercury. If the adhesive force exceeds the cohesive force of the liquid molecules, the angle of contact is very small or almost zero e.g water

Capillarity

It is the tendency of a liquid to rise or fall in a capillarity tube. Cohesion and adhesion as well as surface tension forces are responsible for the capillarity of liquids. In water and soap solution the surface of the liquid or its meniscus curves upward. But in mercury the meniscus is curved downwardsmaway from the glass tube.

Capillarity explains why liquid candle wax rises up the wick of a candle or kerosene rises up the wick of a lamp.

Recommended: Solved questions on linear expansivity

Notes on magnetic field for Cambridge A level and UTME

Magnetic field is a region of space where a magnet or a current carrying conductor or a moving charge experience a magnetic force.

Magnetic field is directed from a North pole to a South pole.

A uniform field, within which the field strength is the same at all points, could be represented as parallel lines that are equally spaced as shown above. The uniform field is stronger if the lines are closer to each other.

Magnetic field around a current carrying wire

When an electric current flow towards you(out of the plane paper), it produces a magnetic field that circulates in anti-clockwise direction. When electric current flow away from you(into the plane paper), it produces field that circulates in a clockwise direction.

Out of the paper is represented by a dot (.)

Into the plane paper is represented by a cross ( x)

Electromagnetism

Force on a current carrying conductor

 

The direction of the force can be detect using Fleming’s left hand rule

• Magnetic Field, B
• Current, I
• Length of conductor in magnetic field, l
• Angle between field and current, θ

For maximum force, θ = 90

F = BIL

B is the magnetic field strength

Magnetic flux density in a magnetic field is defined as the force per unit length acting on a conductor carrying a unit current placed at right angles to the field.

Tesla: The uniform magnetic flux density which, acting normally to a long straight wire carrying a current of 1 amp, causes a force per unit length of 1 Nm^-1 on the conductor

Forces between current-carrying conductors and predict the direction of the forces

When the currents are moving in the same direction, the forces between the current carrying conductor is attractive. When the currrents are moving in opposite direction, the forces is repulsive.

Regardless of the current, the forces on the conductor will be the same.

x is the distance between the two parallel conductors

A moving charge in a magnetic field

F = BQv sinθ

 

For a moving charge in a magnetic field…

 

• Magnetic Field, B
• Charge, q
• Velocity of the charge, v
• Angle between field and velocity of charge,θ

Rectangular ring in a uniform magnetic field

F = NBIL

N is the number of turns pivoted so that it can rotate about a vertical axis

Torque = force x perpendicular distance

Electromagnetic Induction

Faraday noticed that an emf is induced in a circuit whenever there is change in magnetic flux linked with a circuit

Ø = BA

Ø is the magnetic flux and A is the area

Magnetic flux is the product of magnetic flux density and the area normal to the magnetic flux. the unit is weber

NØ  is the magnetic flux linkage

Faraday’s law state that emf induced is directly proportional to the rate of change of magnetic flux linkage.

E α dNØ/dt

Lenz’s law state that the induced emf move in such a way as to oppose the change producing it

E = -NdØ/dt

Lenz’s law is the law of conservation of energy

E = Blv

E is the emf induced

B is the magnetic field strength

l is the length

v is the velocity

Click for answers to some UTME questions on Magnetic field

Notes on ideal gases for Cambridge A level and UTME

An ideal gas is one that obeys the gas laws, and equation of state for ideal gas, at all temperature, pressure and volume. This means Ideal gas obeys

pV = nRT

P = Pressure

V = Volume

T = Temperature

R = universal gas constant

n= number of moles

Infer from a Brownian motion experiment the evidence for the movement of molecules

Brownian motion: random movement of small particles caused be bombardment of invisible molecules

 

• Smoke (oil droplets) are seen to move randomly
• This motion is evidence that the air particles are also moving randomly and colliding with the smoke droplets
• The air particles cannot be seen but their motion can be understood by the smoke droplets which can be seen

Kinetic theory of gases

1. The attraction between molecules is negligible
2. The volume of the molecules is negligible compared with the volume occupied by the gas
3. The molecules are like perfectly elastic spheres
4. The duration of a collision is negligible compared with the time between collisions

Explain how molecular movement causes the pressure exerted by a gas and hence deduce the relationship p = 1/3Nm/V < c 2 >

(N = number of molecules)

• Consider a cube of space with length L
• Consider a particle moving in one dimension x with velocity c_x
• When the particle collides with the wall its velocity is reversed so its change in momentum is equal to…
  • Dp_x = 2mc_x

• The time between collisions with each wall of the cube is equal to…
  • Time between collisions = 2L / c_x

• The rate at which momentum is transferred to the wall is…
  • Rate of change of momentum = 2mc_x / (2L/c_x) = mc_x ² / L

• If there are N particles in the cube the total force is…
  • Total force = Nmc_x ² / L

• Pressure is force over area so pressure is…
  • Pressure on one wall is Nmc_x ² / L³

• L³ is the volume so…
  • Pressure = Nmc_x ² / V

• The average of c_x ² can be written as < c_x ²>
• As all directions, x, y and z can be considered equal
  • < c_x ²> = 1/3< c ²>

• Hence
  • P = 1/3Nm<c ²> / V
• 
• p(rho) = density of gas
• <c²> = mean square speed

It should be carefully noted that the pressure p of the gas depends on the “mean square” of the speed. This is because

1. The momentum change at a wall is proportional to the speed
2. The time interval before this change is repeated is inversely proportional to the speed

Compare pV = 1/ 3 Nm < c 2 > with pV = NkT and hence deduce that the average translational kinetic energy of a molecule is proportional to T.

 

• The average translational E_k of the particles can be expressed as …
  • <E_k> = 1/2m< c²>

• Combining with P = 1/3Nm<c ²> / V we get….
  • pV = 2/3N(1/2m< c²>) = 2/3N<E_k>

• Combining this with pV = NkT we get…
  • pV = 2/3N<E_k> = NkT
  • <E_k> =3/2kT

• Therefore, Temperature is proportional to Average translational kinetic energy

Note

R/N = K

K is the Boltzman’s constant

R is the molar gas constant

Recommended: Solved questions on gravitational field

How to solve questions on capacitance of capacitor for Cambridge A level and UTME

Question 1

Three capacitors, each of capacitance 48 μF, are connected as shown in Fig below

(a) Calculate the total capacitance between points A and B.

(b) The maximum safe potential difference that can be applied across any one capacitor is 6 V. Determine the maximum safe potential difference that can be applied between points A and B.{cambridge A level oct/nov 2014 p43}

solution

for parallel arrangement

c = c1 + c2

c = 48 + 48 = 96uf

96uf is in series with the third 48uf

1/c = 1/c1 + 1/c2

1/c = 1/96 + 1/48

1/c = 3/96

c = 32uf

(b)

In parallel same volatage across the capacitors, while in series same charge across the capacitors

The total charge flowing through the circuit is

C = Q/V

48 x 10^-6 = Q/6

Q = 48*6 x 10^-6 = 288 x 10^-6 c

p.d across the capacitor connected in parallel will be

C = Q/V

96 x 10^-6 = 288 x 10^-6 / V

v = 288 / 96 = 3v

the maximum safe potential difference that can be applied between points A and B = 3 + 6 =9v

or

p.d. across parallel combination is one half p.d. across single capacitor C1
total p.d. = 9 V

Question 2

The combined capacitance between terminals A and B of the arrangement shown in Fig below is 4.0 μF.

Two capacitors each have capacitance C and the remaining capacitors each have capacitance 3.0 μF. The potential difference (p.d.) between terminals A and B is 12 V.
(i) Determine the capacitance C

(ii) Calculate the magnitude of the total positive charge transferred to the arrangement

(iii) Use your answer in (ii) to state the magnitude of the charge on one plate of
1. a capacitor of capacitance C,
2. a capacitor of capacitance 3.0 μF.{cambridge Alevel may/june p42}

solution

(i)

3.0 μF and 3.0 μF are in parallel

c= 3 + 3

c = 6μF

6μF is in series with c and c

1/Ct = 1/6 + 1/c + 1/c

1/4 = 1/6 + 1/c + 1/c

1/4 – 1/6 = 2/c

1/12 = 2/c

C =24uf

(ii)

The potential difference between terminal A and B is 12v

Total capacitance = total charge / total p.d

4 × 10^-6 = Q / 12

Q = 48 × 10^-6 c

(iii)

The charge on capacitor c is the same as the total charge in the arrangement = 48 × 10^-6 c

The charge on 3uf capacitor will be,

Firstly we need to know the p.d across the two capacitors arranged in parallel.

C = Q/v

6uf is the combined capacitance of the two capacitors

6 × 10^-6 = 48 × 10^-6 / v

V = 48/6 = 8v

The charge on the 3uf capacitor will be

3 × 10^-6 = Q/8

Cross multiply

Q = 24 × 10^-6 c

Question 3

The diagram shows three capacitors C1, C2 and C3 of capacitances 2 μF, 6 μF and 3 μF respectively. The potential
differences across C1, C2 and C3 respectively are{UTME 2001}
A. 6V, 2V and 4V.
B. 6V, 4V and 2V.
C. 2V, 6V and 4V.
D. 4V, 6V and 2V.

Solution

The arrangement are in series, so the effective capacitance

1 / c = 1/ C1 + 1/ C2 + 1/ C3

1 / C = 1/ 2 + 1/6 + 1/ 3

1 / C = (3+1+2)/6

1/ C = 1/1

c = 1 μF

note that,

Q = CV

Q = 1*12 =12 C

For series arrangement the same charge will flow through the capacitors

on C1,

12 = 2V

V = 6v

on C2,

12 = 6V

V = 2 v

on C3,

12 = 3V

V = 4 v

The potential difference = 6v, 2v, 4v

A is the correct option

Question 4

The diagram above shows two capacitors P and Q and capacitances 2μF and 4μF respectively connected to
a d.c. source. The ratio of energy stored in P to Q is{UTME 2001}
A. 1 : 2        B. 2 : 1         C. 4 : 1       D. 1 : 4

Solution

The arrangement is parallel, the implication of that is, the same voltage will be supplied across the capacitors

E = 1/2 CV^2

Ep = 1/2 *2*V^2 = V^2

Eq = 1/2*4*V^2 = 2 V^2

Their ratio = Ep /Eq = V^2 / 2V^2 = 1/2

A is the correct answer

Recommended: short note on capacitance of a capacitor

How to solve questions on gravitational field for Cambridge A level and UTME

Question 1

The earth is four times the size of the moon and the acceleration due to gravity on the earth is 80 times that on the moon. The ratio of the mass of the moon to that of the earth is(gravitational force UTME 2004)

A.  1 : 320                            B. 1 : 1280               C. 1 : 80                                 D. 1 : 4

 

Solution

size of earth = 4 x size of moon

r(earth) = 4r(m00n)

G of erath = 80 x g of moon

g = GM/r2

GM(earth)/r2(earth) = 80 *GM(moon)/r2(moon)

M(earth)/16r2(moon) = 80 * M(moon)/r2(moon)

M(earth) = 80*16 M(moon)

M(moon) / M(earth) = 1 / 1280

B is the correct option

Question 2

A binary star consists of two stars A and B that orbit one another, as illustrated in Fig below

The stars are in circular orbits with the centres of both orbits at point P, a distance d from the centre of star A.

i.  The period of the orbit of the stars about point P is 4.0 years. Calculate the angular speed ω of the stars

ii.   The separation of the centres of the stars is 2.8 × 108 km. The mass of star A is MA. The mass of star B is MB.
The ratioMA/MB is 3.0. Determine the distance d.

iii.  Use your answers in (i) and (ii) to determine the mass MB of star B.

May/June 2016 p42

Solution

ii

since they orbit at the same point, there centripetal force will be the same

cross multiply

3d = 2.8 × 108 – d

3d + d = 2.8 × 108

4d = 2.8 × 108

d = 2.8 × 108 / 4

d = 7 x 10^7 km

iii

To calculate for the mass of star B, You must note that the gravitational force will be equal to the centripetal force for star B to orbit star A

Recommended: short note on gravitational field

Capacitance of a capacitor is the charge stored on one plate per unit of potential difference between the plates

c is the capacitance

Q is the charge

v is the potential difference

Farad is the unit of capacitance

 

Uses of capacitor

1. it stores energy
2. smoothing circuit
3. separate charges
4. blocking D.C
5. tuning circuit
6. preventing sparks
7. timing circuit
8. producing electrical oscillations

Explain why the capacitor stores energy but not charge

– capacitor has equal magnitude of +ve and -ve charge

– total charge on capacitor is zero or no resultant charge

– energy stored because there is charge separation

 

What determines the magnitude of capacitance

• distance between the two plates
• Area of overlap of the two plates
• dielectric material

• An isolated metal sphere of radius R has charge +Q on it. The charge may be considered to act as a point charge at the centre of the sphere. Show that the capacitance C of the sphere is given by the expression

substitute the expression for v in c = Q/v

• derive, using the formula C= Q/V , conservation of charge and the addition of p.d.s, formulae for capacitors in series and in parallel

V = V₁ + V₂

V₁ = q / C₁    and   V₂ = q / C₂       (q = charge induced on one plate)

q / C  = q / C₁ + q / C₂

1 / C =  1 / C₁ + 1 / C₂

Parallel arrangement

q = q₁ + q₂

q₁ = C₁V         q₂ = C₂V

CV = C₁V  +  C₂V

C = C₁ + C₂

Energy stored in a capacitor

In other to charge a capacitor work must be done to push electrons onto one plate and off the other. The current stops when the p.d across the capacitor is equal to the e.m.f of the supply. We then say that the capacitor is “fully charged”

Uncharge plates has equal amount of +ve and -ve charge. Connecting the capacitor to supply pulls charge +Q from one plate and transfers it to the other, leaving behind charge -Q. The supply does work in separating the charges. Since the two plates now store equal and opposite charges, the total charge on the capacitor is zero. When we talk about “charge stored” by capacitor, we mean the quantity Q, the magnitude of the charge stored on each plate.

Q = CV

W= E_p = VQ

DE_p = V₀Dq

W = E_p = ½ QV

W = E_p = ½ CV²

Questions and answers to A level and UTME past questions on capacitance of a capacitor

Notes on gravitational field for Cambridge A level and UTME

According to Newton, all masses create a gravitational field in the space round them. The field give rise to a force on any object having mass placed in this field. The moon orbits the earth because it experiences a gravitational force due to the earth’s gravitational field. If an object is placed in a gravitational field, a force will act on the object because of its mass.

What is meant gravitational field?

A region where a mass/body experiences a force of attraction due its mass.

The Earth has a radial field of gravity, which means that the gravitational field is circular and acts from the center point.

The Earths radial gravitational field is represented by the lines

1. The arrows on the field lines show the direction of the gravitational force on a mass placed in the field
2. The spacing of the field lines indicates the strenght of the gravitational field-the farther apart they are, the weaker the field

Gravitational field strenght

Gravitational field strenght at a point is the gravitational force exerted per unit mass on a small object placed at that point.

g = f/m

g = gravitation field strength

m = test mass

units = N kg^-1 = ms^-2

Newton’s law of universal gravitation: it states that any two point masses attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of their separation

G is the gravitational constant 6.67 x 10^-7 Nkg^-2m²

since f= GMm/r2

g =f/m = GMm/mr2 = GM/r2 (This equation does not depend upon the mass of the small object )

g is the gravitational field strenght

Gravitational potential energy: at a point is the work done in bringing a mass from infinity to a point

gravitational potential is always negative: because of the force of attraction that exist between two masses

• Zero of potential energy is at infinity
• Potential energy taken as a negative value
• The work done in moving a mass between two points in a gravitational field is independent of the path taken

Centripetal Acceleration
•For an orbiting satellite, the gravity provides centripetal force which keeps it in orbit

Geostationary Orbits

•Geostationary satellite is one which is always above a certain point on the Earth
•For a geostationary orbit: T = 24 hrs. and orbital radius is
a fixed value from the center of the Earth

By using r³ / T² the radius of orbit needed for geostationary orbit can be calculated

Recommended: Questions and answers on gravitational field

How to solve questions on electric field for Cambridge A level and UTME

The answers below give you a step by step on how to solve questions on electric field when writing Cambridge A level physics. Follow the steps and you can comment if you have any question on electric field.

Question 1

Two parallel metal plates, 4.0 cm apart, are at electric potentials of 800 V and 2000 V. Points X, Y and Z are situated in the space between the plates at distances of 1.0 cm, 2.0 cm and 3.0 cm from the lower plate

What is the electric field strength, in V m–1, at X, Y and Z?

{Cambridge A level may/june 2016 p12}

Solution

Electric field strenght at any point in two parallel plate is constant i.e it is a uniform electric field strenght

E = V/d

V is the potential difference between the plate = 2000 – 800 = 1200 v

d is the distance between the plate = 4.0 cm = 0.04 m

E = 1200 / 0.04 = 30000 vm-1 = 3 x 10^4 vm-1

C is the correct answer

You can download physics textbook for free that I have prepared for students writing Cambridge A level, UTME, and ost UTME by Clicking here

Question 2

Two parallel vertical metal plates are connected to a power supply, as shown in Figure below

An α-particle travels in a vacuum between the two plates. The electric field does work on the α-particle. The gain in kinetic energy of the α-particle is 15 keV. Calculate the electric field strength between the plates

{Cambridge A level 2016 may/june p22}

Solution

workdone = gain in kinetic energy

workdone = force x distance between the plates

workdone = qE x d

The charge on alpha particle = 2e

workdone = qEd = 3.2 x 10^-19 x E x 16 x 10^-3

wordone = 51.2 x 10^-22 E

1eV = 1.6 x 10^-19 J

1KeV = 1.6 x 10 ^-16 J

15 KeV = 24 x 10^-16 J

51.2 x 10^-22 E = 24 x 10^-16 J

E = 24 x 10^-16 / 51.2 x 10^-22

E = 0.46875 x 10 ^6 Vm-1

E = 4.7 x 10^5 Vm-1

Question 3

An oil droplet has charge –q and is situated between two horizontal metal plates as shown in the diagram below.

The separation of the plates is d. The droplet is observed to be stationary when the upper plate is at potential +V and the lower plate is at potential –V.
For this to occur, what is the weight of the droplet?{Cambridge A level 2015 may/june p11}

Solution

The droplet is stationary because;

the weight on the droplet = the electric force on the droplet

The electric force = qE

E = potential difference / distance between the plate

Potential difference = V – (-V) = 2V

E = 2V / d

Electric force = 2Vq / d

Weight on the droplet = 2Vq/d

B is the correct option

Recommended: Short note on electric field