2. IX. Electricity
· Electrical circuits: basic elements.
· Current.
· Potential difference.
· Resistance.
· Ohm's law.
· Effects of electrical resistors in series and in parallel.
· Effects of electricity: chemical, thermal, magnetic
and mechanical.
· Joule Effect.
· Electrical power.

3. Charge and current
The charge of electricity is measured in the unit coulomb (C).
There is a link between the charge and current:
If the charge flows at this rate: Then the current is:
1 coulomb per second 1 ampere
2 coulombs per second 2 amperes
etc.
The equation is therefore the following:
Charge = current x time
(C) = (A) x (s)

4. An electrical cell makes electrons (e-) move. An
example of an electrical cell is a battery.
Current
A flow of charge is called a current. A charge is
the flow of electrons through matter.
The SI unit to measure the strength of a current
is ampere (A)
More or less 6.000.000.000.000.000.000 e- flow
round a circuit per second to give a strength of 1
A.
For smaller currents milliamperes (mA) are used
as a unit. 1000mA = 1A

5. Current direction
Most circuit diagrams, like the one below, have an arrow
indicating the direction of the current flow. The flow always
goes from positive to negative round the circuit.
As electrons are negatively charged they are repelled by the
negatively charged side of the cell and go therefore in the
opposite direction as the current flow.

6. The potential differece (PD) or voltage
The scientific name for voltage is potential difference (PD)
The higher the voltage the more energy a cell gives to the electrons it
pushes out.
The PD is measured in volt (V).
If the PD accross a cell is 1 volt, then 1 joule of potential energy
are given to each coulomb of charge. So 1 volt = 1 joule per
coulomb 1V=(j/C)
To produce a
higher PD, cells
can be put in
series.
Cells in parallel and in series

7. When batteries are connected in series you
have to sum up their voltage.
The voltage tells you how much energy is
given per Coulomb.
So if you connect batteries in series, the
voltage given is the sum of all.

8. Copy this table

9. Resistance
The resistance of a material indicates how well it is able to prevent the
electrons from flowing through it.
The resistance is calculated with the following
equation:
Resistance ( ) =
Ω
Current through conductor (A)
PD accros the conductor (V)
So the lower the resistance, the less PD is needed to
give the same current
For example if a PD of 6 V is needed to make a
current of 3A flow through a wire the resistance
would be: 6V/3A= 2Ω
1 kilohm (kΩ) = 1000 Ω
1 megaohm (MΩ) = 1.000.000 Ω

10. Ohm’s law
1. Calculate the resistance of a conductor through
which a current of 2 A flows with a PD of 12V.
Sol. 6Ω
2. Calculate the PD between the extremes of a
conductor with a resistance of 10 Ω and a current of
7.5 A. Sol. 75 V
3. Calculate the intensity of a current which circulates
through a conductor with a resistance of 10 Ω, if
the PD between its extremes 0.02 mV.
Sol. 2·10-6
A

11. Series and Parallel circuits
Bulbs in series
·The bulbs share the PD (V)
from the battery, bulbs glow
dimly
·If one bulb would be removed,
the would go out because the
circuit is broken.
Bulbs in parallel
·Each gets the full PD from the battery because they are both
connected directly to it.
·If one bulb would be removed teh other would keep on working
becasue the circuit isn't broken

12. Circuit in series
Intensity/Current (A)
IT = I1 = I2 = I3 etc
PD (V)
VT = V1 + V2 + V3 etc
Resistance (Ω)
RT= R1 + R2 + R3 etc
Circuits in parallel
Intensity/Current (A)
IT = I1 + I2 + I3 etc
PD (V)
VT = V1 = V2 = V3 etc
Resistance (Ω)
1 = 1 + 1 + 1 etc
RT R1 R2 R3
1 = 1 + 1 + 1 etc
RT 4 5 6
Or:
0.25 + 0.20 + 0.16 = 0.61
So:
1 = 1.64Ω
0.61
For
example

13. 9)
a) The combined resistance.
b) The current which flows through the circuit.
c) The voltage in each resistance.
Sol. a) 23 Ω; b) Sol. 0.52 A;
c) V1= 7.8V, V2 = 2.6V, V3 = 1.6V
10. Given the following circuit, calculate:
a) The combined resistance of the circuit.
b) The current that circulated through the circuit.
Sol. a) 1.6 Ω; b) 7.5 A

15. Power
Power = energy transformed
time taken
W= Joules/Seconds
For electrical energy we use the unit kwh (Kilowatt per hour)
1 kw= 1000 watts
1kwh= 1Kw per h
or kwh= kw
h

16. Electrical Power
Power = PD x current
Watt = Volt x Ampere
or
P= VI
So for a battery:
Power = 12V x 2A = 24W
For bulb A: 8V x 2A = 16W
For bulb B: 4V x 2A = 8W

17. 4.- Calculate the energy consumed by an iron of
1000 W if it is switched on during 30 minutes. Sol.
0.5 kwh
5.- We have a light bulb of 75 W and 220 V.
Calculate a) the current that passes through the bulb; b)
the resistance of the bulb; c) the energy consumed
during 30 days if it will be switched on for 4 hours a day.
Sol. a)

18. 1. Calculate the energy consumed by an iron of 1000 W if
it is plugged to a 220 V socket during 30 minutes.
Sol. 0.5 kwh
2. We have a light bulb of 75 W and 220 V. Calculate a)
the current that passes through the bulb; b) the
resistance of the bulb; c) the energy consumed during
30 days if it will be switched on for 4 hours a day.
Sol. a) 0.34 A; b) 647 Ω; c) 9 kwh.
3. Calculate how much energy it will cost to dry your
hair every day if you use a hair dryer of 1100 W during 20
minutes each. The Price per KWh is €0.15. Sol. 0.055 €
4. Calculate the amount of energy consumed and its price
per month if you use a 350 W television during 3 hours
each day. The Price per KWh is €0.15.
Sol. 31,5 kwh/mes - 4.72 €.

19. Circuits 1
Circuits 2
Resistor
resistors-in-parallel
ohms-law_en.jar
circuit-construction-kit-dc_en.jar

21. Adjuntos
circuit-construction-kit-dc_en.jar
ohms-law_en.jar