TOPIC 2: CURRENT ELECTRICITY – PHYSICS NOTES FORM 2

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CURRENT ELECTRICITY

Current electricity is the rate at which an electricity source will make charges to flow or pass a certain point in a conductor or in an electric circuit.

This means that, when electrical devices are joined in an electric circuit, electrons flow in a continuous path. It is the flow rate of which is referred to as current electricity or electric current.

The sources of electricity are of different nature which include;
  1. Generators
  2. Charged capacitors
  3. Dry cells
  4. Dynamo
  5. Solar panels

The device in which energy transformation occurs and is maintained is called electrical source.

Load – is the device used for transforming the electrical energy into any of the sensible forms of energy. for example; heat, light, mechanical or sound energy etc.

The quantity of electricity is measured in coulombs denoted by ‘C’.

It follows that, if the quantity of charge of 6 coulombs is passed at a certain point of a conductor in 3 seconds, then;

C:\thlb\cr\tz\__i__images__i__\h2.jpgElectric current = 2 coulombs/second or 2C/s

SIMPLE ELECTRIC CIRCUITS

Electric circuit – is a continuous path formed by connecting electrical devices. Such as Battery, switch, socket etc along which electrons can flow.

A simple electric circuit may consist of;

  1. Cell or any source of current electricity.
  2. Switch or control to allow or restrict the flow of current electricity.
  3. Conductor to transmit the current electricity.
  4. Load to consume the supplied power delivered by the current electricity.

The figure below shows the simple electric circuit containing a sound (cell), a control (switch) and the load (bulb);

C:\thlb\cr\tz\__i__images__i__\hh1.jpg

The electric devices used in a circuit are called circuit, components or elements.

USES OF COMPONENTS OF ELECTRIC CIRCUIT

1. Cell or battery – Is a source of electric current.

2. Switch – Is a device used to switch on an electric current in order to allow the flow of an electric current or off an electric current thus to stop the flow of the electric current respectively.

3. Resistor – Is a component included in an electric circuit because of its resistance to current electricity flow. There are variable value resistors and fixed value resistors, all are made of resistance wire or carbon.

4. Ammeter – Is an instrument with low internal resistance used for measuring electric current.

5. Connecting wire – Is a material used to provide a direct path which allows the flow of current between two points in a circuit and are used to connect circuit components.

6. Capacitor – Is an electric conductor or a system of electrical conductors which can store electric charge.

7. Voltmeter – Is an instrument with high internal resistance used for measuring potential difference of any two points in an electric circuit.

8. Socket and plugs – Are devices that connect electrical appliances to the power supply so that electric power can flow through them.

CONCEPT OF CURRENT, VOLTAGE AND RESISTANCE IN AN ELECTRIC CIRCUIT

The electric current flows from a point with low potential to the point with high potential.

The S.I unit of electric current is Ampere denoted by ‘A’. The electric current (I) encounters resistance (R) along its path, which will result in voltage drop (V) in a circuit.

The S.I unit of resistance is Ohms denoted as Ω.

The potential difference (P.d) which causes electric current to flow is defined as “the energy per coulomb consumed when electricity moves from one point to another”.

The S.I unit of potential difference is volt (V). This voltage is given as the product of current flowing between two points and resistance offered between the two points

I.e. Voltage = Current In amperes X Resistance in Ohms

V = I x R

 V = IR 

OHM’S LAW

-It states that “the voltage across the conductor is directly proportional to the electric current flowing if temperature is constant”

C:\thlb\cr\tz\__i__images__i__\HHH1.jpg

The resultants shows that the resistance of a wire (P.d) is proportional to the current flowing through the conductor

I.e. Vα I

Introducing proportionality constant ‘K’

V = KI

The constant of proportionality is called the resistance ‘R’ thus,

V = IR

COMBINATION OF RESISTORS

Resistors can be connected either in series or parallel depending on the magnitude of effective resistance required. Series connection gives a bigger value of effective resistance and the parallel connection gives small value of effective resistance.

RESISTORS IN SERIES

By connecting resistors in series, when the switch ‘S’ is closed, the current ‘I’ which flows through the circuit flows through each resistor.
C:\thlb\cr\tz\__i__images__i__\HHHH.jpg

Total resistance between points A and B which is commonly referred to as equivalent resistance (Req) will produce a potential difference in the circuit given by ohm’s law as;

V=IReq

The voltage across each resistor in the circuit is given by V1 =IR1 and V2 =IR2

The sum of the voltage drops equal to the potential difference in the circuit (i.e. potential difference between (A and B)

Total voltage = V1 + V2

... VT = V1 + V2

Total voltage = V1 + V2

Since V = IR, V1 = IR1 and V2 = IR2

IRT = IR1 + IR2

IRT =I (R1 + R2 )

RT = R1 + R2

RESISTORS IN PARALLEL

In the figure below I is the current in the main circuit. On the other hand I1 and I2 are current through individual resistors R1 and R2.

The sum of all currents through the resistors which are connected in parallel gives the value of current equal to the main circuit.
C:\thlb\cr\tz\__i__images__i__\HHHHH1.jpgTherefore, IT = I1 + I2

If RT is the equivalent resistance of the main circuit between A and B, then by Ohm’s law the current is given by;

C:\thlb\cr\tz\__i__images__i__\i1.jpg

From IT = I1 + I2

C:\thlb\cr\tz\__i__images__i__\ii1.jpgOn diving both sides by V

C:\thlb\cr\tz\__i__images__i__\iii2.jpgCross multiplication

1(R1R2) = RT (R1 + R2)
C:\thlb\cr\tz\__i__images__i__\iiii1.jpg

For two resistors connected in parallel.

EXAMPLES

1.Given that R1= 4Ω and R2= 6Ω, find the equivalent resistance when the resistors are connected.

  1. In parallel

2.   In series

Solution

1.Series

C:\thlb\cr\tz\__i__images__i__\j1.jpg
RT = R1+R2

RT = 4Ω + 6Ω

RT = 10Ω

2.Parallel

C:\thlb\cr\tz\__i__images__i__\jj3.jpg
C:\thlb\cr\tz\__i__images__i__\jjj1.jpg

= 2.4Ω

2.Two conductors of resistance 4Ω and 5Ω are connected in series across a 60V supply. Find;

  1. The total resistance
  2. The current in the circuit
  3. The potential difference across each resistor

C:\thlb\cr\tz\__i__images__i__\k1.jpg RT = R1+R2

= 4Ω + 5Ω

= 9Ω
the total resistance = 9Ω

C:\thlb\cr\tz\__i__images__i__\kk2.jpg

I = 6.7A
Potential difference across R1

V1 = IR1

V1 = 6.7 x 4 = 26.8v

Potential difference across R2

V2 = IR2

V2 = 6.7 x 5 = 33.5v

Total current = 26.8 + 33.5 = 60A

3.Consider the circuit shown below. What will be the reading on the Ammeter?

Solution

V = 12V

C:\thlb\cr\tz\__i__images__i__\kkk2.jpg

C:\thlb\cr\tz\__i__images__i__\KKKK3.jpgRT =2Ω

C:\thlb\cr\tz\__i__images__i__\l2.jpg

I = 6A

EXERCISE

1. In a circuit, the amount of charges passing through a point is 9 coulombs in 4.5 seconds. What is the electric current passing at that point?

Solution

Quantity of charges = 9 coulombs

Time = 4.5 sec

Electric current =?

C:\thlb\cr\tz\__i__images__i__\ll3.jpg

Electric current = 2coulombs/sec

2. The two resistances 15Ω and 5Ω are connected in series across 20v supply, find;

  1. Total resistance
  2. The total current in the circuit
  3. The current through each resistor

Solution

Data given

R1 = 15Ω

R2 = 5Ω

Voltage = 20v

The total resistance

RT = R1 + R2

= 15Ω + 5Ω

= 20Ω

The total current in a circuit (I)

From V = IR
C:\thlb\cr\tz\__i__images__i__\LLL2.jpg

But v = 20v, R = 20Ω

C:\thlb\cr\tz\__i__images__i__\LLLL1.jpg
 I = 1A

The current through each resistor

C:\thlb\cr\tz\__i__images__i__\LLLLL1.jpg

But V = 20V, R1 =15ΩC:\thlb\cr\tz\__i__images__i__\M1.jpg

= 1.3A

C:\thlb\cr\tz\__i__images__i__\MM1.jpgbut V = 20 V, R = 5Ω

C:\thlb\cr\tz\__i__images__i__\MMM1.jpg   I2 = 4A

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