# BMAT Physics Notes: Electricity

Advice & Insight From BMAT Specialists

## â€‹Electrostatics

Atoms are normally electricallyÂ neutral.
The total negative charge isÂ equalÂ to the total positive charge.
Â
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 â€˜charging by frictionâ€™.
Â
The direction of the electron flow depends on the materials used.
â€‹
If the clothÂ losesÂ electrons then it becomesÂ positivelyÂ charged; the rodÂ gainsÂ electrons and becomesÂ negativelyÂ Â charged.

If the electrons flow in theÂ oppositeÂ direction theÂ  materials gain theÂ oppositeÂ charge (as shown in the second diagram).

ConductorsÂ Â 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 calledÂ â€˜earthingâ€™.

Experiment shows that objects that have theÂ sameÂ chargeÂ repelÂ and objects that haveÂ oppositeÂ chargesÂ attract,Â Â we sayÂ â€˜like charges repel, unlike charges attractâ€™.

[NOTE:Â NeverÂ answer electrostatic questions in terms of â€˜moving protonsâ€™ or â€˜moving positive chargesâ€™. Although some books talk about a â€˜build up of positive chargesâ€™ this isÂ neverÂ Â 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 itÂ losesÂ electrons].

## â€‹Electric CURRENT (I)

When a battery or cell is connected to a conductor (e.g. a metal) anÂ electric currentÂ will flow. This is due to a net movement ofÂ Â negative charges. The â€˜chargeâ€™ is carried byÂ electronsÂ that move through the metal.
[REMEMBER:Â 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].
Â
Definition: The CURRENT is the rate of flow of charge.
Â
HenceÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â CurrentÂ Â Â  =Â Â Â Quantity of charge
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  time
â€‹or,Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â IÂ  =Â  Q/tÂ Â Â Â  orÂ  QÂ  = I tÂ Â Â
Â
whereÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Q = charge inÂ coulomb (C)Â Â Â Â Â Â IÂ  = current in Amps (Amperes)Â (A)
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â tÂ  = inÂ seconds (s)
Â
[From this it follows thatÂ Â 1 CÂ  = 1 A.s ]Â Â
[You willÂ NOTÂ be asked to recall the definitions BUT you will be expected toÂ use the equations in calculations].

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##### Examples 1
1.Â  How muchÂ chargeÂ flows past a point when:
Â (a) 3A flows for 4sÂ  __________________________________
Â (b) 20 mA flows for 1 minuteÂ  _________________________
Â (c) 500 mA flows for 10 minutes ________________________
Â
2.Â  Find theÂ currentÂ that flows ifÂ
Â (a) 10 C of charge pass a point in 5 s _____________________Â
Â (b) 1000 C pass a point in 4 minutes _____________________Â
Â (c) 12kC of charge pass a point in 10 minutes ______________Â

3.Â Â Find theÂ timeÂ taken when Â
Â (a) A current of 5A causes a charge of 20 kC to pass __________________ Â
Â (b) A current of 10mA causes a charge of 20 C to pass __________________ Â
â€‹ (c) A current of 25 ÂµA causes a charge of 20 mC to pass __________________

## d.c./a.c.

When a battery is connected in a circuit the current flowsÂ one wayÂ (from theÂ positiveÂ terminal around the circuit to theÂ negativeÂ terminal). This is calledÂ direct current (d.c.)
Â
If a dynamo is connected to a circuit then the current keepsÂ changing direction.Â This is calledÂ alternating current (a.c.)

## Voltage

Whenever current flows through a component the electronsÂ lose energyÂ (usually in the form ofÂ heat).

The amount of energy transferred whenÂ oneÂ coulomb of charge passes through a component is called theÂ voltageÂ across the component. This leads to an important equation:

Â VoltageÂ  =Â Â Â energyÂ Â Â Â  orÂ Â Â Â Â  V =Â  Â Â EÂ
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  chargeÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Q
â€‹

• 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 calledÂ potential difference (p.d.)

Learn time-efficient BMAT strategies and practice with reflective BMAT questions & worked solutions.

##### â€‹Voltage and Energy Calculations 2

â€‹1.Â  (a) A charge of 20 C passes through a load and does 50 J of work. Calculate the p.d. across the load.
Â  Â  Â Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Â

Â  Â  Â (b) 100J of heat are produced whenÂ  5 C of charge pass through a resistor. Calculate the voltageÂ across the resistor.
Â  Â  Â _________________________________________
Â
Â
2.Â  A current of 2A flows through a lamp for 10 minutes.
Â  Â  Â If 12 kJ of work is done calculate

Â  Â  Â (a)Â the amount of charge that passedÂ Â  _______________________ Â
Â  Â  Â (b)Â the p.d. across the lampÂ Â  _______________________Â Â
Â
3.Â Â A current of 100mA flows through a lamp for 5 minutes. If the p.d. across the lamp is 240 V calculate:

Â  Â  Â (a)Â The quantity of charge that passesÂ  _____________________ Â
Â  Â  Â (b) The work done by the chargeÂ  _________________________

## Voltmeters

Voltmeters are used to measureÂ voltage. They are connected inÂ parallelÂ across a component.

AnÂ idealÂ voltmeter is one with anÂ infiniteÂ resistance. This meansÂ noÂ current flows through it when it is connected across a component and so doesÂ notÂ alter the voltage it is measuring.

## Ammeters

Ammeters are used to measureÂ current.Â They are connected inÂ seriesÂ with components.
Â
AnÂ idealÂ ammeter is one withÂ zeroÂ resistance. This means that when they are connected in circuits there isÂ no changeÂ in current that it is measuring.

## â€‹Resistance

Def:Â The resistance of a material is the ratio of p.d. to current.
Â
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ResistanceÂ  =Â  voltage /currentÂ Â Â Â  orÂ Â Â Â Â  RÂ  =Â Â VÂ Â Â  (In texts it is usually written:Â  V = I RÂ Â
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  I
Â
[You will need to be able to use this equation to find V, I or R.]

##### Unit of Resistance
WhenÂ VÂ is measure in VOLTS andÂ IÂ in AMPS then resistance is inÂ OHMS (Î©)

â€‹Good conductors (e.g. metals) have aÂ lowÂ electrical resistance.
Poor conductors (insulators) have aÂ highÂ electrical resistance (such as rubber, plastic).

##### â€‹Examples 3

1.Â  Calculate the p.d. when Â
Â  Â  (a)Â A current of 6A flows through a resistor of 2.5 Î©,Â  Â  ______________________________Â Â
Â  Â  (b)Â A current of 20mA flows through a resistor of 1.8kÎ©, Â ____________________________
Â  Â  (c)Â A current of 50mA flows through a resistor of 0.5MÎ©, Â ____________________________
â€‹
2.Â  Calculate theÂ currentÂ when Â
Â  Â  (a)Â A p.d. of 10V is applied across a resistor of 2.5Î©,Â  Â ______________________________Â Â Â Â
Â  Â  (b)Â A p.d. of 6V is applied across a resistor of 2 kÎ©,Â Â Â Â Â  _____________________________
â€‹Â  Â  (c)Â A p.d of 12V is applied across a resistor of 0.1MÎ©,Â  Â _____________________________

3.Â  Calculate theÂ resistanceÂ of a component if Â
Â  Â  (a)Â A p.d. of 12V produces a current of 0.5A, Â __________________________________
Â  Â  (b)Â A p.d. of 6V produces a current of 3 mA, Â __________________________________
Â  Â  (c) A p.d of 230V produces a current of 10ÂµAÂ  _________________________________

## Thermistors

AÂ thermistorÂ is a component whoseÂ resistanceÂ depends on itsÂ temperature.
Its resistance could increase or decrease with increasing temperature. We only deal with resistance that decreasesÂ withÂ increasing temperature.
[Such a material is said to have aÂ negative temperature coefficient of resistanceÂ  (NTC.)]
In theÂ COLDÂ its resistance isÂ HIGHÂ and whenÂ HOTÂ its resistance isÂ LOW

The graph shows how the resistance of a thermistor varies with temperature.
The resistance can vary from a few hundred ohms whenÂ coldÂ to a fewÂ  ohms whenÂ hot.

## Light-Dependent-Resistor (LDR)

A light dependent resistor is a component whoseÂ resistanceÂ depends on theÂ light intensityÂ falling on it.
â€‹
In theÂ DARKÂ its resistance isÂ HIGHÂ and in theÂ LIGHTÂ its resistance isÂ LOW
The resistance can vary from a few million ohms (MÎ©) to a few hundred.Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Â
The diagram shows the variation of resistance with light intensity for a typical LDR.

## The Diode

A diode is a component that only allows current to flowÂ one wayÂ through it. It allows current to flow in the direction of the â€˜triangle.â€™
It is used to protect devices from incorrect power connections.
â€‹
The diagram shows the I-V characteristics for a typical diode.
Â
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 theÂ resistanceÂ of the diode must beÂ decreasing.
Â
Note that below about 0.5V the current is ZERO.
Â
This means that the resistance of the diode below 0.5V isÂ  _________________

## Filament Lamp

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 filamentÂ increases.
This increase in temperature causes the resistance of the filament to INCREASE.
This means that a graph of I against V isÂ non-linear

## â€‹Use of Ammeters in circuits

In the diagram shown if ammeters are used to measure the currents it is found that

In general,
Â
TOTAL CURRENT flowing INTO a junction = TOTAL CURRENT flowing OUT.

##### Example 4
Calculate the magnitude and direction of the current I shown in the diagram.

## Series

In the diagram shown the current I is the same through both resistors.
â€‹Hence it follows

So, the general result:
Â In a SERIES circuit the CURRENT is the SAME at ALL points.

## â€‹Parallel

Since the total current flowing IN = total current flowing OUT of a junction:

## â€‹Voltmeters in series and parallel

Series

In theÂ  circuit shown:

Parallel

In the parallel circuit shownÂ Â V1Â  =Â  V2

This isÂ ALWAYS TRUEÂ Â no matter what the resistor values are.

##### â€‹Formulae for Resistors in SERIES and PARALLEL

SERIES

When two (or more) resistors are connected in SERIES the total resistance is the SUM of the resistors.

PARALLEL

When two (or more) resistors are connected in PARALLEL, the total resistance is given by:

where RTÂ  is the TOTAL resistance (sometimes called EQUIVALENT resistance).
Â
[NOTE that the value of RT is ALWAYSÂ LESSÂ than either value of each resistor]

Example 5

Calculate the total resistance when two resistors of 3Î©Â and 6Î©Â are connected in (i) series, (ii) parallel.

## â€‹Short circuit

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 acrossÂ eitherÂ 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 â€˜short circuitâ€™.

##### â€‹Example 6

In the circuit shown, the ammeter reading isÂ IÂ and the voltmeter isÂ V

When the switch is closed, which row describes what happens toÂ IÂ andÂ VÂ ?

## â€‹Power & Energy

POWERÂ is defined as the amount of ENERGY transferred PER SECOND.
Â
This leads to aÂ generalÂ power equation:Â Â Â

â€‹ForÂ electricalÂ circuits it can be shown that:Â  Power = voltage x currentÂ  orÂ Â Â PÂ  = V I
Â
SinceÂ  V = I R we can substitute for V and rewrite the equation:Â  Â  Â

â€‹ENERGYÂ can be calculated by rearranging the â€˜powerâ€™ equation:Â  Energy = power x time orÂ Â E = V I t
[Unit: J] [ The timeÂ tÂ MUST be inÂ seconds].
You need to be able to use the equations in calculations.

##### Example 7

A lamp is rated at â€˜100W, 230Vâ€™.Â  Calculate
Â
Â  (a) TheÂ currentÂ flowing,
Â  (b) TheÂ resistanceÂ of the lamp,
Â  (c) TheÂ energyÂ transferred if the lamp is switched on for 1 hour.

Examples 1
1. (a) Â 12CÂ Â Â Â (b) 20Â  1.2CÂ Â Â Â Â  (c) 0.3CÂ

2. (a) 2AÂ  Â (b) 4.17AÂ Â Â  (c) 20AÂ

3. (a)Â  4000sÂ  (b)Â Â  2000sÂ Â Â Â  (c)Â Â  800sÂ

Examples 2
1. (a) 2.5VÂ Â Â  (b) 20VÂ Â Â Â
2. (a) 1200 CÂ Â Â  (b)Â  10V
3. (a) 30 CÂ Â Â Â Â  (b) 7200J
â€‹
Examples 3Â
1. (a) 15VÂ Â Â  (b) Â 36VÂ Â Â  (c)Â  25VÂ Â
Â Â Â
2. (a)Â  4AÂ  (b) 3mAÂ Â  (c)Â  0.12 mA

3. (a)Â  24WÂ Â Â Â  (b) 2000WÂ Â Â Â  (c) Â 23MWÂ
Â
â€‹
Example 4:Â Â  infinite
Â

Example 5:Â Â 5.5A outwards
Â

Example 6:Â  9W.,Â  2W
Â
â€‹
Example 7
When the switch is closed the parallel resistor is â€˜short circuitedâ€™. 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).
Â
0.43AÂ Â Â  (b)Â  529WÂ Â  (c)Â  3.6 x 105 J

#### BMAT Physics Notes: Electricity

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