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Activity 1
To be able to explain the most important measuring units and their mutual connection
LO 2.4  
We have already said that an electrical current is the flow of electrons in a circle. The speed at which the electrons flow, is measured in Ampère. In order to measure amperage, one must allow the current to flow through an Ammeter.
The battery supplies the energy that is also called the electromotive force (EMF ). It is measured in Volts . In order to measure voltage one must place the two points of a voltmeter on either side of a component.
Resistance is offered against the flow of the current and the resistance is measured in Ohms .
A battery does not provide its full energy because some of its energy is lost. This is called power and it is measured in Watts .
Ohm’s law states that the current through a resistor is directly proportional to the tension that is coupled over a resistor, i.e. the current is directly proportional to the tension and inversely proportional to the resistor. It is formulated as follows:
I = V / R
Actuation is the energy that must supply the battery and is wasted as heat in a resistor. It is formulated as follows:
P = V x I
Symbol  Unit  Abbreviation  
TENSION  V  Volt  V 
CURRENT  I  Ampère  A 
RESISTANCE  R  Ohm  Ώ 
POWER  P  Watt  W 
The formulas make it possible for one to determine units through mathematical computations, e.g. if one has the Volts and Ampère of a circuit, one can calculate the power. Nowadays we use advanced apparatus to measure current, tension and resistance.
Assignment 1
1. Provide two more formulas that can be derived from the formula I = V / R.
2. Provide two more formulas that can be derived from the formula P = V x I.
LO 2.4  
Learning outcomes(LOs) 
LO 2 
TECHNOLOGICAL KNOWLEDGE AND UNDERSTANDING The learner will be able to understand and apply relevant technological knowledge ethically and responsibly. 
This is demonstrated when the learner: 
structures:2.1 demonstrate knowledge and understanding of frame structures:

Ohm's law:
Assignment 1
1. V = I × R and R = V/I
2. V = P/I en I = P/V
3. R = V/I = $\frac{3}{\mathrm{1,5}}$ = 2 ohm
4. P = V × I = 3V × 1,5A = 4,5 watt
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