BMAT Physics Notes: Electricity
Advice & Insight From BMAT Specialists
Atoms are normally electrically neutral.
The total negative charge is equal to the total positive charge.
However, it is very easy to move electrons from an atom.
For example if an insulating rod is rubbed with a suitable cloth it is possible for electrons to move from surface atoms as shown.
This is called ‘charging by friction’.
The direction of the electron flow depends on the materials used.
If the cloth loses electrons then it becomes positively charged; the rod gains electrons and becomes negatively charged.
If the electrons flow in the opposite direction the materials gain the opposite charge (as shown in the second diagram).
Conductors cannot be charged because the charge will flow through the material (to earth).
A build up of charge on a material can be dangerous. A spark may occur which could ignite any combustible (e.g. petrol vapour) material nearby.
For this reason materials are usually connected to the ground by a conductor so that the charge can flow to (or from) the ground. This prevents a build up of charge.
This is called ‘earthing’.
Experiment shows that objects that have the same charge repel and objects that have opposite charges attract, we say ‘like charges repel, unlike charges attract’.
[NOTE: Never answer electrostatic questions in terms of ‘moving protons’ or ‘moving positive charges’. Although some books talk about a ‘build up of positive charges’ this is never due to moving positive charges or protons. Protons, which are positively charged, are held strongly in the nucleus and cannot be moved easily. A material ALWAYS becomes positive charged because it loses electrons].
Electric CURRENT (I)
When a battery or cell is connected to a conductor (e.g. a metal) an electric current will flow. This is due to a net movement of negative charges. The ‘charge’ is carried by electrons that move through the metal.
[REMEMBER: Although protons are positively charged they DO NOT move through any material. This is because they are held by very strong forces in the nucleus of an atom].
Definition: The CURRENT is the rate of flow of charge.
Hence Current = Quantity of charge
or, I = Q/t or Q = I t
where Q = charge in coulomb (C) I = current in Amps (Amperes) (A)
t = in seconds (s)
[From this it follows that 1 C = 1 A.s ]
[You will NOT be asked to recall the definitions BUT you will be expected to use the equations in calculations].
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(a) 3A flows for 4s __________________________________
(b) 20 mA flows for 1 minute _________________________
(c) 500 mA flows for 10 minutes ________________________
2. Find the current that flows if
(a) 10 C of charge pass a point in 5 s _____________________
(b) 1000 C pass a point in 4 minutes _____________________
(c) 12kC of charge pass a point in 10 minutes ______________
3. Find the time taken when
(a) A current of 5A causes a charge of 20 kC to pass __________________
(b) A current of 10mA causes a charge of 20 C to pass __________________
(c) A current of 25 µA causes a charge of 20 mC to pass __________________
When a battery is connected in a circuit the current flows one way (from the positive terminal around the circuit to the negative terminal). This is called direct current (d.c.)
If a dynamo is connected to a circuit then the current keeps changing direction. This is called alternating current (a.c.)
Whenever current flows through a component the electrons lose energy (usually in the form of heat).
The amount of energy transferred when one coulomb of charge passes through a component is called the voltage across the component. This leads to an important equation:
Voltage = energy or V = E
- You need to be able to use this equation to find V, E or Q;
- Also note that V must be in VOLTS; E in JOULES and Q in COULOMBS.
- Voltage is often called potential difference (p.d.)
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Voltage and Energy Calculations 2
1. (a) A charge of 20 C passes through a load and does 50 J of work. Calculate the p.d. across the load.
(b) 100J of heat are produced when 5 C of charge pass through a resistor. Calculate the voltage across the resistor.
2. A current of 2A flows through a lamp for 10 minutes.
If 12 kJ of work is done calculate
(a) the amount of charge that passed _______________________
(b) the p.d. across the lamp _______________________
3. A current of 100mA flows through a lamp for 5 minutes. If the p.d. across the lamp is 240 V calculate:
(a) The quantity of charge that passes _____________________
(b) The work done by the charge _________________________
Voltmeters are used to measure voltage. They are connected in parallel across a component.
An ideal voltmeter is one with an infinite resistance. This means no current flows through it when it is connected across a component and so does not alter the voltage it is measuring.
Ammeters are used to measure current. They are connected in series with components.
An ideal ammeter is one with zero resistance. This means that when they are connected in circuits there is no change in current that it is measuring.
Def: The resistance of a material is the ratio of p.d. to current.
Resistance = voltage /current or R = V (In texts it is usually written: V = I R
[You will need to be able to use this equation to find V, I or R.]
Unit of Resistance
Good conductors (e.g. metals) have a low electrical resistance.
Poor conductors (insulators) have a high electrical resistance (such as rubber, plastic).
1. Calculate the p.d. when
(a) A current of 6A flows through a resistor of 2.5 Ω, ______________________________
(b) A current of 20mA flows through a resistor of 1.8kΩ, ____________________________
(c) A current of 50mA flows through a resistor of 0.5MΩ, ____________________________
2. Calculate the current when
(a) A p.d. of 10V is applied across a resistor of 2.5Ω, ______________________________
(b) A p.d. of 6V is applied across a resistor of 2 kΩ, _____________________________
(c) A p.d of 12V is applied across a resistor of 0.1MΩ, _____________________________
3. Calculate the resistance of a component if
(a) A p.d. of 12V produces a current of 0.5A, __________________________________
(b) A p.d. of 6V produces a current of 3 mA, __________________________________
(c) A p.d of 230V produces a current of 10µA _________________________________
A thermistor is a component whose resistance depends on its temperature.
Its resistance could increase or decrease with increasing temperature. We only deal with resistance that decreases with increasing temperature.
[Such a material is said to have a negative temperature coefficient of resistance (NTC.)]
In the COLD its resistance is HIGH and when HOT its resistance is LOW
The graph shows how the resistance of a thermistor varies with temperature.
The resistance can vary from a few hundred ohms when cold to a few ohms when hot.
A light dependent resistor is a component whose resistance depends on the light intensity falling on it.
In the DARK its resistance is HIGH and in the LIGHT its resistance is LOW
The resistance can vary from a few million ohms (MΩ) to a few hundred.
The diagram shows the variation of resistance with light intensity for a typical LDR.
A diode is a component that only allows current to flow one way through it. It allows current to flow in the direction of the ‘triangle.’
It is used to protect devices from incorrect power connections.
The diagram shows the I-V characteristics for a typical diode.
It starts to conduct when the p.d. across it is about 0.5V and from thereafter the current rises quite steeply. This means that the resistance of the diode must be decreasing.
Note that below about 0.5V the current is ZERO.
This means that the resistance of the diode below 0.5V is _________________
Inside a lamp the filament is a coil of metallic wire.
When COLD the filament has a LOW resistance.
When the lamp is switched on the temperature of the filament increases.
This increase in temperature causes the resistance of the filament to INCREASE.
This means that a graph of I against V is non-linear
Use of Ammeters in circuits
TOTAL CURRENT flowing INTO a junction = TOTAL CURRENT flowing OUT.
Hence it follows
So, the general result:
In a SERIES circuit the CURRENT is the SAME at ALL points.
Voltmeters in series and parallel
In the parallel circuit shown V1 = V2
This is ALWAYS TRUE no matter what the resistor values are.
Formulae for Resistors in SERIES and PARALLEL
When two (or more) resistors are connected in SERIES the total resistance is the SUM of the resistors.
When two (or more) resistors are connected in PARALLEL, the total resistance is given by:
where RT is the TOTAL resistance (sometimes called EQUIVALENT resistance).
[NOTE that the value of RT is ALWAYS LESS than either value of each resistor]
Calculate the total resistance when two resistors of 3Ω and 6Ω are connected in (i) series, (ii) parallel.
In the circuit shown current will flow through both resistors so the ammeter A and voltmeter V will both show readings.
If the wire is placed across either resistor then ALL the current will flow through the wire and NONE through the resistors. BOTH the ammeter and voltmeter will read ZERO.
This is sometimes called a ‘short circuit’.
In the circuit shown, the ammeter reading is I and the voltmeter is V
When the switch is closed, which row describes what happens to I and V ?
Power & Energy
POWER is defined as the amount of ENERGY transferred PER SECOND.
This leads to a general power equation:
For electrical circuits it can be shown that: Power = voltage x current or P = V I
Since V = I R we can substitute for V and rewrite the equation:
ENERGY can be calculated by rearranging the ‘power’ equation: Energy = power x time or E = V I t
[Unit: J] [ The time t MUST be in seconds].
You need to be able to use the equations in calculations.
A lamp is rated at ‘100W, 230V’. Calculate
(a) The current flowing,
(b) The resistance of the lamp,
(c) The energy transferred if the lamp is switched on for 1 hour.
- (a) 12C (b) 20 1.2C (c) 0.3C
- (a) 2A (b) 4.17A (c) 20A
- (a) 4000s (b) 2000s (c) 800s
- (a) 2.5V (b) 20V
- (a) 1200 C (b) 10V
- (a) 30 C (b) 7200J
- (a) 15V (b) 36V (c) 25V
- (a) 4A (b) 3mA (c) 0.12 mA
- (a) 24W (b) 2000W (c) 23MW
Example 4: infinite
Example 5: 5.5A outwards
Example 6: 9W., 2W
When the switch is closed the parallel resistor is ‘short circuited’. NO current flows through it. ALL the current flows through the switch. The voltmeter reads ZERO. The TOTAL circuit resistance DECREASES so the current INCREASES. So the answer is (B).
0.43A (b) 529W (c) 3.6 x 105 J
BMAT Physics Notes: Electricity
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