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2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } - Education (3) - Nairaland

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Nairaland 2016 Jamb Tutorial Classroon [use Of English Thread] / Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] / Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] (2) (3) (4)

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 8:29am On Dec 24, 2015
iceberryose:
Sir when's our next class? Jamb is fast approaching sir
Tomorrow after today sir. We still have a lot of topics to go round.
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 8:59am On Dec 24, 2015
[b]Electrostatics.
Experimental discovery shows that whenever two bodies are rubbed together, they acquire attracting property on some light objects such as paper. Consider one ebonite rod rubbed with a fur; they are seen to attract each other. The two bodies are said to be electrified. Other examples are glass and silk.
The law of electrostatic states that like charges repel each other and unlike charges attract each other.
Electrostatic Induction.
Electrostatic induction is a method of charging a body by introducing a charged body to a neutral body. When this is done, the body is said to be induced with a charge.
Note; To charge a body positively, a negatively charged rod is brought near and to charge a body negatively, a positively charged rod is brought near.
Gold Leaf Electroscope.
Gold leaf electroscope is an instrument for measuring, indicating and detecting charges. The tin foil diverge when there is a potential difference between the leaves; this process measures the charge in the electroscope.
Distribution of Charges on Conductors.
Experimental works have shown that charges are distributed where there is a sharp curve. The density of these charges are greater at the surface of sharp curve. The charge per unit area of a charged surface is called surface density.
Surface density is greater at the corner or pointed edge than at the plain surface.
lightning Conductor.
Lightning conductor is used to prevent a tall building from being damaged when it is struck by lightning. The conductor is a long metal rod installed or connected to the Earth by means of a cable. The sharp outer point of the top gains an induced charge opposite to that in the thunder cloud. The charged ionizes the nearby air and the charged air molecules flow upwards from the point. This discharges the cloud before a lightning flash occurs.[/b]

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 9:29am On Dec 24, 2015
[b]Capacitor.
A capacitor is a device for storing electrical charges. It consists of two conductors which are parallel to each other.
The storage of a capacitor is increased due to an increase in its conductivity.
Function of Capacitor.
1. Separate ac from dc.
2. Control current in ac.
3. Storing of electrical charge or energy.
Arrangement of a Capacitor.
A capacitor is arranged in either a series or in a parallel arrangement.
Parallel Arrangement.
---||---c1
---||---c2
---||---c3
c = c1 + c2 + c3.
Hence, the capacitance of capacitors in parallel is the sum of their individual capacitance.
Series Arrangement.
---||---||---||---
---c1---c2---c3---.
1/c = 1/c1 + 1/c2 + 1/c3.
Hence, the capacitance of capacitor in series is the sum of their reciprocal.

Example; Two capacitors of capacitance 3uF and 6uF are connected in series. Calculate the equivalent capacitance.
Solution;
c1 = 3uF,
c2 = 6uF.
Capacitors connected in series
1/c = 1/c1 + 1/c2.
C=2uF.

Types of Capacitor.
1. Variable air capacitor; This type of capacitor is commonly used in electric circuit.
2. Paper capacitor; This sis a parallel plate capacitor.
3. Mica capacitor; This is used where many capacitors are required.[/b]

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 9:31am On Dec 24, 2015
i'm so sorry for not being very active here. It is well embarassed
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 10:02am On Dec 24, 2015
[b]Capacitance.
The capacitance of a capacitor is defined as the ratio of electric charge to the potential difference between or across the plates of the capacitors.
It is denoted by C, measured in Farad(F) and a scaler quantity, thus, charge in any plate is directly proportional to the potential difference across the capacitor.
Mathematically;
charge ~ potential difference
Q ~ V
Q = CV
where c is constant called capacitance.
Factors Affecting Capacitors.
1. The capacitance of a capacitor is directly proportional to the area.
C ~ A.
2. Capacitance is inversely proportional to the distance between the two plates (di-electric).
C ~ 1/D.
3. Capacitance is directly proportional to relative permitivity.
C ~ E.
Mathematically;
C ~ AE/D
C = AE/D if constant is 1.
Recall that,
Q = CV, then,
C = Q/V, therefore,
Q/V = AE/D.
Note; Dielectric is an insulating medium which separates two metal plates eg air, mica etc.
Energy Stored by a Capacitor.
The workdone is given by;
W = average pd X charge.
W = (1/2)QV.

Example; A capacitor has a capacitance of 10uF and the charge on the plate is 10-5C. Calculate the energy stored by the capacitor.
Solution;
Energy = 5 X 10-6J.[/b]
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 10:15am On Dec 24, 2015
Classwork.
1. Find the capacitance of a paper capacitor whose distance between the plate is 5cm and area 10mm2, where E = 8.85 X 10-12F/m.
2. The energy stored in a capacitor of capacitance 5uF is 40J. Calculate the voltage applied across its terminals.
Cc Debbychris Iceberryose Geofavor Oxytocin
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 10:17am On Dec 24, 2015
Geofavor:
i'm so sorry for not being very active here. It is well embarassed
No lele sir, "we are all waiting for you".
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by iceberryose(m): 12:30pm On Dec 24, 2015
HOPE AM RIGHT
WOULD HAVE SUBMITTED EARLIER BUT SMALL COLD ENTER BODY.
1.C=? A=10mm=1cm D=5cm E=8.85 x 10^-12
using C=AE/D C=1 x 8.85 x 10^-12/5 c=1.77 x 10^-12

2. W=40j c=5uF V=?

W=1/2 x QV W=QV/2
RECALL That Q=CV
W=cv[v]/2 40=5v(v)/2 80=5v^2 v^2=16 v=4volts
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 7:42am On Dec 26, 2015
iceberryose:
HOPE AM RIGHT

WOULD HAVE SUBMITTED EARLIER BUT SMALL COLD ENTER BODY.

1.C=?
A=10mm=1cm
D=5cm
E=8.85 x 10^-12

using C=AE/D
C=1 x 8.85 x 10^-12/5
c=1.77 x 10^-12


2. W=40j
c=5uF
V=?


W=1/2 x QV
W=QV/2

RECALL That Q=CV

W=cv[v]/2
40=5v(v)/2
80=5v^2
v^2=16
v=4volts
Great job, nicely done. Lets talk about conversion first.
1. To convert from mm to cm, "we" divide by 10.
2. To convert from cm to m, "we" divide by 100.
3. Therefore; to convert from mm to m, "we" divide by 1000.
4. To convert from mm2 to cm2, "we" divide by 10 and another 10, why? It is beacuse of the square.
5. To convert from cm2 to m2, "we" divide by 100 and another 100, why? Same as above.
6. To convert from mm2 to m2, "we" divide by 1000 and another 1000, why? Same as above.
7. The standard unit of measurement of length is m not mm.
Bonus; To convert mm4 to cm4, "we" divide by 10, and 10, and 10 and 10. Why? The power is 4.
Can I assume you get the logic? This principle works for all units in power.
Merry Christmas and Happy New Year in Advance.

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 7:51am On Dec 26, 2015
thankyouJesus:
1. Find the capacitance of a paper capacitor whose distance between the plate is 5cm and area 10mm2, where E = 8.85 X 10-12F/m.
d=5cm=0.5m.
A=10mm2=0.00001m2.
E=8.85 X 10-12F/m.
C=AE/d
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 7:56am On Dec 26, 2015
thankyouJesus:
2. The energy stored in a capacitor of capacitance 5uF is 40J. Calculate the voltage applied across its terminals.
E=40J.
C=5uF=5 X 10-6F (u is a scientific notation).
W=(1/2)QV.
Recall that,
Q=CV.
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 9:44am On Dec 28, 2015
Thankyoujesus, well done sir. pls i'd like to know how you arrived at 5 x 10-6joules in that example.
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 5:06pm On Dec 28, 2015
thankyouJesus:
Example; A capacitor has a capacitance of 10uF and the charge on the plate is 10-5C. Calculate the energy stored by the capacitor.
Solution;
Energy = 5 X 10-6J.[/b]
c=10uF=10X10-6F=10-5F.
Q=10-5C.
E=(1/2)QV.
Recall that Q=CV.
V=Q/C.
E=(1/2)Q2/C.
Geofavor:
Thankyoujesus, well done sir. pls i'd like to know how you arrived at 5 x 10-6joules in that example.
thankyouJesus:
Example; A capacitor has a capacitance of 10uF and the charge on the plate is 10-5C. Calculate the energy stored by the capacitor.
Solution;
Energy = 5 X 10-6J.[/b]
c=10uF=10X10-6F=10-5F.
Q=10-5C.
E=(1/2)QV.
Recall that Q=CV.
V=Q/C.
E=(1/2)Q2/C.

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 8:06am On Dec 29, 2015
Good morning all, compliment of the season.
Electric Current.
Electric current through a conductor is the rate of flow of charges (electrons).
The unit of current is ampere (Amp, A). It is denoted by I and is a vector quantity. Electric current is measured with the aid of an ammeter.
The continuous flow of current is known as electricity.
Mathematically;
I = Q/t
Q = It.
Where Q is charge (C),
I is current (A),
t is time (s).

Example; A 20uc flows through a conductor in 10ns. Determine the current flowing in the circuit.
Solution;
Q=20uc=20X10-6c.
T=10ns=10X10-9s.
I=Q/t
I=2X103A.

Effects of Electric Current.
1. Magnetic effect eg electromagnet.
2. Heating effect eg iron.
3. Chemical effect eg charging battery.

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 8:33am On Dec 29, 2015
[b]Potential Difference.
Potential difference is the amount of workdone in moving a unit positive charge from a point to another, mostly from a point of lower potential to the point of higher potential. It is denoted by p.d and measured in volts(v) with the aid of a voltmeter.
Mathematically;
V = W/Q.
W = QV.
Where W is workdone (J).
Q is charge (C).
V is potential difference (v).

Example; A 50uc charge flows through a conductor. If the p.d between it ends is 20mv, find the workdone by the charge.
Solution;
Q=50uc=50X10-6c.
V=20mv=20X10-3v.
W=QV
W=1X10-6J.
Voltage is the potential difference between two points in a given electric circuit. It is denoted by v and measured in volts(v).
Electric Circuit.
Electric circuit is a representation of how current moves from the source of the current (eg battery or cell) through resistor and other devices before entering the source.
There are two types of electrical circuit, the open and the closed circuit.
1. Open circuit; A circuit is said to be open if the source of electricity is not connected to any external conductor.
-------------
| |
| |
-----|¡------.
2. Closed circuit; A circuit is said to be closed if the source of electricity is connected to an external conductor through which current is passed.
__
------| |----||---/ ----
| --- |
| |
| |
-----------|i------------.
[/b]
cc; fynestboi good morning o

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 8:50am On Dec 29, 2015
thankyoujesus, complements of the season to you too. More power to your elbow.

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 9:14am On Dec 29, 2015
[b]Resistance.
Resistance is the amount of opposition given to the flow of electric current through a conductor of electricity. It is denoted by R, measured in Ohms.
Mathematically;
R = V/I.
V = IR.
Factors affecting resistance.
1. Temperature.
2. Cross-sectional area.
3. Length of conductor.
4. Resistivity.
Measurement of resistance.
The methods of measuring resistance are;
1. Voltmeter-ammeter method.
2. Wheatstone bridge.
3. Metre bridge.
4. Potentiometer.
1. Voltmeter-ammeter method; This method is through the use of graph by adjusting or varying the rheostat, different values of the voltage will be obtained.
Slope(resistance)=change in V/change in I.
2. Wheatstone bridge; This is an instrument used in getting accurate resistance. This consists of four resistors connected side by side to form a close circuit.
R1/R2=R3/R4.
3. Metre bridge; This is a bridge with low temperature co-efficient and length 100cm or 1m.
Example; The balance length on a metre bridge for two resistances P and Q are 50cm and 60cm respectively. If P is 10 Ohms, calculate the value of Q.
Solution;
R1/R2=L1/L2.
10/Q=50/60
Q=12 Ohms.
4. Potentiometer; This is a device used to measure potential difference. It consists of uniform wire of length 100cm stretched on a wooden board by the side of a metre rule. The wire is at a low temperature co-efficient.
Example; The balance length of a potentiometer wire for a cell of emf 1.62V is 90cm. If the cell is replaced by another one of emf 1.08V, calculate its new balance.
Solution;
E1/E2=L1/L2.
1.62/1.08=90/x
x=60cm.[/b]

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 9:46am On Dec 29, 2015
[b]Resistance in Series.
Resistances are in series when two or more resistance are connected end to end, so that the same current flows through all of them, though they have different voltages.
-----[]----[]---[]-----
R=R1 + R2 + R3.
Resistance in Parallel.
----[]----
----[]----
----[]----.
1/R = 1/R1 + 1/R2 + 1/R3.
Ohm's Law.
It states that the electric current in a given metallic conductor is directly proportional to the potential difference applied between its ends, provided that temperature and other physical factors are constant.
Conductors that obey Ohm's law are called Ohmic conductors eg metals and alloys.
Conductors which do not obey Ohm's law are called non-ohmic conductors eg divide value, thermostat, etc.
Resistivity.
This is defined as the resistance per unit length per unit cross-sectional area of the material concerned. It is denoted by P and measured in Ohm/m.
Mathematically;
R=PL/A.
Where R is resistance (Ohm).
P is resistivity (Ohm/m).
L is length (m).
A is area (m2).
Electrical Conductivity.
This is the reciprocal of resistivity. It is denoted by K and measured in m/Ohm.
Mathematically;
K = 1/P.
Where P is resistivity.[/b]

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 9:54am On Dec 29, 2015
Note;
1. Energy stored by a capacitor,
W=(1/2)QV.
Amount of workdone in moving charge,
W=QV.
2. Capacitors in parallel,
C = C + C + C.
Resistors in parallel,
1/R = 1/R +1/R + 1/R.
3. Capacitors in series,
1/C = 1/C + 1/C + 1/C.
Resistors in series,
R = R + R + R.

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 10:13am On Dec 29, 2015
Classwork.
1. Write short note on (a) galvanometer (b) voltmeter (c) ammeter.
2. The resistivity of a given wire of cross sectional area 1X107mm2 is 20 Ohm/m. Calculate (a) the resistance of the wire if the length is 200cm (b) electrical conductivity. (Hint; watch out for the units).
3. What is the combined resistance of 5 Ohm wire arranged in series with a parallel arrangement of 4 Ohm and 6 Ohm wires?
4. A potentiometer wire carrying a steady current is 1m long. With a standard cell of emf 1.1V, a suitable balanced length of 440cm was obtained. What is the emf of a cell which gives a balance length of 680cm?
5. A wire 100cm long and of diameter 6000mm has a resistance of 2 Ohm. What is the resistivity of the material of which it is made? (Hint; Area of circle).
Cc Debbychris Iceberryose Geofavor Oxytocin.
Starting from next year, I should be able to make reference of JAMB past questions, please bear with me.
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 12:29pm On Dec 29, 2015
thankyouJesus:
[b]Classwork.
1. Write short note on (a) galvanometer (b) voltmeter (c) ammeter.
a) a galvanometer is used for detecting very little amount of current. It can detect current as low as the miliamperes.

b) a voltmeter is the device used for measuring the emf and the potential difference in an electric circuit.

C) the ammeter is used for measuring the current flowing in an electric circuit in Amperes.


2. The resistivity of a given wire of cross sectional area 1X107mm2 is 20 Ohm/m. Calculate (a) the resistance of the wire if the length is 200cm (b) electrical conductivity. (Hint; watch out for the units).
a) R = pl/A
R = (20 x 200 x 10-2)./(1 x 107. x 10-3).
R = 40/104.
R = 40/104.
R = 40/10000
R = 0.004ohms

b) K = 1/p
K = 1/20
K = 0.05(m)ohm-1 grin


3. What is the combined resistance of 5 Ohm wire arranged in series with a parallel arrangement of 4 Ohm and 6 Ohm wires?

Rs. = 5ohms.

1/Rp. = 1/4 + 1/6.

Rp. = 2.4ohms.

: . RT. = 5 + 2.4 = 7.4ohms.


4. A potentiometer wire carrying a steady current is 1m long. With a standard cell of emf 1.1 Ohm, a suitable balanced length of 440cm was obtained. What is the emf of a cell which gives a balance length of 680cm?
..... I think you meant 1.1V

so, E1./E2. = L1./L2.

1.1V/E = 440cm/680cm

1.1V/E = 4.4m/6.8m

E = (1.1V x 6.8 )/4.4

E = ....1.7V....


5. A wire 100cm long and of diameter 6000mm has a resistance of 2 Ohm. What is the resistivity of the material of which it is made? (Hint; Area of circle).
firstly, R = pl/A, pl = RA, p = RA/l
R = 2ohms
l = 100cm = 1m
A = 3.143m x (6000/2)2mm.
= 3.143m x 30002mm.
= 3.143m x 32m.
= 28.287m2.

. : p = (2 x 28.287)/1
= 56.574ohm/m


Starting from next year, I should be able to make reference of JAMB past questions, please bear with me.[/b]
no frorblem sir. cool
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by iceberryose(m): 1:41pm On Dec 29, 2015
thankyouJesus:
Classwork.
1. Write short note on (a) galvanometer (b) voltmeter (c) ammeter.
2. The resistivity of a given wire of cross sectional area 1X107mm2 is 20 Ohm/m. Calculate (a) the resistance of the wire if the length is 200cm (b) electrical conductivity. (Hint; watch out for the units).
3. What is the combined resistance of 5 Ohm wire arranged in series with a parallel arrangement of 4 Ohm and 6 Ohm wires?
4. A potentiometer wire carrying a steady current is 1m long. With a standard cell of emf 1.1 Ohm, a suitable balanced length of 440cm was obtained. What is the emf of a cell which gives a balance length of 680cm?
5. A wire 100cm long and of diameter 6000mm has a resistance of 2 Ohm. What is the resistivity of the material of which it is made? (Hint; Area of circle).
Cc Debbychris Iceberryose Geofavor Oxytocin.
Starting from next year, I should be able to make reference of JAMB past questions, please bear with me.
Lemme go to number 2.


2. A=1x10^7mm^2
=1000m^2
P=20 ohm/m
L=20m


P=RA/L
20=RX1000/20
400=1000R
R=0.4Ohms


K=1/P
K=1/20
K=0.05m/ohms


3. Series
R=5 ohms

Parallel
1/R=1/4 + 1/6
1/R=5/12

5R=12
R=2.4 ohms

R=5 + 2.4
R=7.4ohms


4. E1=1.10emf
E2=?
L1=44m
L2=68m


E1/E2=L1/L2
1.1/e=44/68
74.8=44e
e=1.7emf


5. L=10m,d=6m,radius=3m,R=2 Ohms,P=?


A=pir^2
A=pi x 3^2
A=pi x 9
A=22/7 x 9
A=198/7
A=28.26

P=RA/L
P=2 X 28.26/10
P=5.652ohm/m
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 2:58pm On Dec 29, 2015
thankyouJesus:
Great job, nicely done. Lets talk about conversion first.
1. To convert from mm to cm, "we" divide by 10.
2. To convert from cm to m, "we" divide by 100.
3. Therefore; to convert from mm to m, "we" divide by 1000.
4. To convert from mm2 to cm2, "we" divide by 10 and another 10, why? It is beacuse of the square.
5. To convert from cm2 to m2, "we" divide by 100 and another 100, why? Same as above.
6. To convert from mm2 to m2, "we" divide by 1000 and another 1000, why? Same as above.
7. The standard unit of measurement of length is m not mm.
Bonus; To convert mm4 to cm4, "we" divide by 10, and 10, and 10 and 10. Why? The power is 4.
Can I assume you get the logic? This principle works for all units in power.
Merry Christmas and Happy New Year in Advance.
cc Geofavor Iceberryose, both of you need to go over this again and make necessary corrections where applicable. Nice job by the way.
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by DebbyChris(f): 8:15pm On Dec 29, 2015
Nawa, you guys have gone far ooo, i have been busy, i will try my best to be available.
thankyoujesus, thank you for the good work and God bless you.. prosperous new year in advance
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 9:52pm On Dec 29, 2015
thankyouJesus:
both of you need to go over this again and make necessary corrections where applicable. Nice job by the way.
pls highlight the mistakes i made in converting. i have checked again, but couldnt spot any, except one little arithmetic error.
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 9:52pm On Dec 29, 2015
DebbyChris:
Nawa, you guys have gone far ooo, i have been busy, i will try my best to be available.
thankyoujesus, thank you for the good work and God bless you.. prosperous new year in advance
God bless you too my sister, prosperous new year in advance.
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 9:55pm On Dec 29, 2015
Geofavor:

pls highlight the mistakes i made in converting. i have checked again, but couldnt spot any, except one little arithmetic error.
Check your area conversion in number 2.
Cc Ibrokane
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by Geofavor(m): 9:58pm On Dec 29, 2015
thankyouJesus:
Check your area conversion in number 2.
Cc Ibrokane
just seeing it now. Thanks

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Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 12:45pm On Dec 30, 2015
thankyouJesus:
2. The resistivity of a given wire of cross sectional area 1X107mm2 is 20 Ohm/m. Calculate (a) the resistance of the wire if the length is 200cm (b) electrical conductivity.[/b]
Area = 1X107mm2 = 1X10m2.
P =20 Ohm/m.
L = 200cm = 2m.
R = PL/A.
Cc Iceberryose Geofavor
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 12:49pm On Dec 30, 2015
iceberryose:
5. L=10m,d=6m,radius=3m,R=2 Ohms,P=?


A=pir^2
A=pi x 3^2
A=pi x 9
A=22/7 x 9
A=198/7
A=28.26

P=RA/L
P=2 X 28.26/10
P=5.652ohm/m
You need to correct the conversion in red, great work by the way.
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 1:10pm On Dec 30, 2015
[b]Good afternoon all.
Electrical Energy.
Electrical energy is the workdone when a quantity of charge moves between two points of potential differences measured in Joule.
Mathematically;
W = QV;
Recall that Q = It.
W = IVt-----------(1).
Recall that V = IR.
W = I2Rt-------(2).
W = V2t/R -----(3).

Example; Calculate the electrical energy produced by a heater with a voltage supply of 220V, when a current of 10amps passed through it for 5minutes.
Solution;
I = 10amps.
V = 220V.
T = 5 X 60 = 300sec.
W = 660000J.

Electrical Power.
Electrical power is the rate at which the energy is used up. Power is measured in watts with wattmeter.
Mathematically,
Power = workdone/time
P = IVt/t.
P = IV ------------(1).
P = I2R ---------(2).
P = V2/R.

Example; Calculate the power dissipated by a heater of 220V and a resistance of 10 Ohms.
Solution;
V = 220V.
R = 10 Ohms.
P = 4840watts.[/b]
Re: 2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } by thankyouJesus(m): 1:27pm On Dec 30, 2015
Types of Power.
1. Real Power; This is the power used in driving machines in the factories in which (cos¤) power factor is considered.
Mathematically;
Real power = IVcos¤.
2. Apparent Power; This is the power available for any electrical appliance which is simply structured.
Mathematically;
Apparent power = IV.
Note; Power factor = Real power/ Apparent power.

Example; An electric welding machine has a power factor of 0.5 when the apparent power is 100w, find the real power.
Solution;
Power factor = 0.5
Apparent power = 100w.
Power factor = Real power/Apparent power.
Real power = 50w.

The other types of power are active and reactive power, they are not in JAMB syllabus

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