Page 200 - WCS - Electrical

WORKSHOP CALCULATION & SCIENCE - CITS

11 R = 40 ohms
1
R = 60 ohms
2
V = 220 v
I = ______ A

I = ______ A
1
I = ______ A
2

12 R = ______ W
2
R = ______ W

Ohm’s law, relation between V.I.R & related problems
Ohm’s law
V - Voltage in volts

l - Current in Ampere
R - Resistance in ohms.
In any closed circuit the basic parametres of electricity (Voltage, Current and resistance) are in a fixed relationship
to each other.
Basic values
To clarify the basic electrical values, they can be compared to a water tap under pressure
Basic electricity - Ohm’s law, relation between V.I.R & related problems
Water pressure Electron pressure Voltage
Exercise 1.7.35
Amount of water Electron flow Current
Ohm’s law Throttling of tap Obstruction to Resistance

V - Voltage in volts
l - Current in Ampere
R - Resistance in ohms.
In any closed circuit the basic parametres of electricity
(Voltage, Current and resistance) are in a fixed relationship
to each other.
Basic values

To clarify the basic electrical values, they can be compared
to a water tap under pressure
Water pressure - electron pressure - Voltage Relationships
Relationships
Amount of water - electron flow - Current If the resistance is kept constant and the voltage is
throttling of tap - obstruction to - Resistance increased, the current is increased
If the resistance is kept constant and the voltage is increased, the current is increased
electron flow I  V
If voltage is constant and the resistance is increased, current is decreased
If voltage is constant and the resistance is increased,
current is decreased

l 

Ohm’s law
187
From the above two relationships we obtain Ohm’s law,
CITS : WCS - Electrical - Exercise 18
which is conveniently written as V = R.I.

Ohm’s law states that at constant temperature
the current passing through a closed circuit is
directly proportional to the potential differ-
ence, and inversely proportional to the resis-
tance.

By Ohm’s law

EXAMPLE
A bulb takes a current of 0.2 amps at a voltage of 3.6 volts.
Determine the resistance of the filament of the bulb to find
R. Given that V = 3.6 V and l = 0.2 A.

To find ‘R’. Given that V = 3.6V and I = 0.2 A
Therefore V = l x R
3.6 V = 0.2 A x R

Therefore

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