Electric circuitsGrade 11

CurrentA source of energy is needed for an electric current to flow.

ddd dd

coulomb

ddd dd

coulomb

SourceA cell is a source of electrical energy for charges.

Current

Current is the flow of charge.

Charge is a collective noun for billions of charges. We measure charge in coulomb.

++++++

+++++++++++

+

++++++

++++++

++++++-

--

-

-

- --

-

-

-

-

--

-

-

-

- --

-

-

-

-

--

-

-

-

-

--

-

-

-

--

--

-

-

-

--

-

-

-

- --

-

-

-

-

--

-

-

-

- --

-

-

-

-

--

-

-

-

-

--

-

-

-

--

--

-

-

-

Conventional currentThe direction for the current is from the positive pole of the cell, through the circuit to the negative pole of the cell.

ddd dd

coulomb

ddd dd

coulomb

1V

Current

Current (I) is the rate of flow of charge (Q), measured in ampère.

I =𝑄

∆𝑡

Source of energyCharges transfer the energy obtained from the cells through the circuit.

ddd dd

coulomb

ddd dd

coulomb

1V

EMF

The emf of a cell or battery is the total amount of energy a cell can transfer to one coulomb of charge.

Potential difference

Potential difference (V) is the amount of work (W) done per coulomb of charge to move it between two points in a circuit, measured in volt.

V = 𝑊

𝑄

ddd dd

coulomb

The Voltmeter

The potential difference across a cell is 1 volt, when 1 joule of energy is transferred to each coulomb of charge.

ddd dd

coulomb

1V

ddd dd

coulomb

The Voltmeter

The potential difference across a cell is 2 volt, when 2 joule of energy is transferred to each coulomb of charge.

ddd dd

coulomb

2V

ddd dd

coulomb

The VoltmeterThe potential difference across the light bulb is 1 volt, when 1 joule of energy is transferred to the light bulb by each coulomb of charge.

ddd dd

coulomb

ddd dd

coulomb

The VoltmeterThe potential difference across the light bulb is 2 volt, when 2 joule of energy is transferred to the light bulb by each coulomb of charge.

ddd dd

coulomb

ddd dd

coulomb

The VoltmeterThe potential difference across each light bulb is 1 volt, because 1 joule of energyis transferred to each light bulb by each coulomb of charge. The potentialdifference across both light bulbs together is 2 volt, because 2 joule of energy is transferred by each coulomb of charge while moving through both of the bulbs.

ddd dd

coulomb

ddd dd

coulomb

2dd

d dd

coulomb

The VoltmeterThe voltmeter must be connected in parallel.

ddd dd

coulomb

ddd dd

coulomb 1

1

2dd

d dd

coulomb

The VoltmeterThe voltmeter has a very high resistance and no current flows through it.

ddd dd

coulomb

ddd dd

coulomb 1

2

0

ddd dd

coulomb

The VoltmeterThe main current of 2 ampère (2 coulomb per second) splits and each coulomb of charge gains 1 joule of energy when it moves through one of the cells. Therefore the potential difference across the cells is 1 volt.

ddd dd

coulomb

1V

ddd dd

coulomb

ddd dd

coulomb

1V

Practical Investigation

Determine the relationship between the potential difference (V) across a resistor and the current (I) through the resistor at a constant temperature.

A

V

R

R

I

VR =

Graph of V versus IV

I1

2

3

4

5

6

1 2 3 4 5 60

Ohm’s law

The current through a resistor is directly proportional to the potential difference across its ends if the temperature remains constant.

V ∝ I

Ohmic conductorObeys Ohm’s law

E.g. nichrome wire

Non-ohmic conductorDoes not obey Ohm’s law

E.g. light bulb, LED

The gradient,

V/I, remains

the same

here and is

called the

resistance.

V

I

V ∝ I

Resistance

Resistance (R) is the relationship between the potential difference across the ends of a conductor and the current flowing through the conductor.

R = 𝑉

𝐼

Resistance

A resistor offers resistance against the flow of charge.

ddd dd

coulomb

ddd dd

coulomb

R1 R2 R3

VTV1 V2

V3A1

A2

I1 = I2

VT = V1 + V2 + V3

Series:

Rs = R1 + R2 + R3

The more resistors in series,

the greater the resistance

and the less the current.

R1 R2 R3

VTV1 V2

V3A1

A2

V1 = V2 = V3 = VT

IT = I1 + I2 + I3

R2

R1

I3

I2

I1

VT

V1

V2

V3321

1111

RRRRp

++=

Parallel:

R₃

The more resistors

in parallel, the less

the resistance and

the greater the

current.

R

R2

R1

I3

I2

I1

VT

V1

V2

V3

R₃

Homework

p. 296,no. 8

p. 308,nos. 8, 10, 13, 14

Electrical energy (W) in circuits

W = V.QW = V.I.ΔtW = I².R.Δt

W = 𝑉2.∆𝑡

𝑅

measured in joule (J)

Power

Power (P) is the rate at which electrical energy is dissipated, measured in watt (W).

P = 𝑊

∆𝑡

Also P = V.I = I².R = 𝑉²

𝑅

Cost of electricity

P = 𝑊∆𝑡

W = P.Δt

number of units = power (kW) x time (h)

Cost = number of units x price/unit

Homework

p. 321,nos. 3, 4, 13, 16, 17, 19, 20