Through experimentation, Georg Simon Ohm established a mathematical relationship between resistance, current intensity and potential difference.
Ohm’s law is represented by the following equation:
|V = R\times I|
where
|V:| potential difference |\text {(V)}|
|R:| resistance |(\Omega)|
|I:| current intensity |\text {(A)}|
There are three applications of Ohm’s law:
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If the current intensity increases, the potential difference increases.
-
If the resistance increases, the potential difference increases.
-
If the resistance increases, the current intensity decreases.
To determine the resistance of a component in an electrical circuit, a simple electrical circuit must be created with that component. So, a circuit that contains only that component and a power supply is needed. An ammeter and a voltmeter will be connected to this circuit to measure the current intensity based on the potential difference.
Here are the measurements obtained for the resistor in the above circuit.
|
Potential difference |\text {(V)}| |
Current intensity |\text {(A)}| |
| |0| | |0| |
| |1| | |0.0005| |
| |2| | |0.0010| |
| |3| | |0.0015| |
| |4| | |0.0020| |
| |5| | |0.0025| |
| |6| | |0.0030| |
Using the lab data, the graph of the potential difference in relation to the current intensity gives the relationship below.
The slope of this graph is the resistance |\text {(R)}.|
||\begin{align}R=\dfrac{V_{2}-V_{1}}{I_{2}-I_{1}} \quad \Rightarrow \quad R &= \dfrac{6\ \text {V} - 0\ \text{V}}{0.0030\ \text {A} - 0\ \text{A}} \\ R &= 2\ 000 \ \Omega \end{align}||
What is the resistance of a filament in a |6\ \text{V}| lamp with an electrical current of |250 \ \text{mA}| flowing through it?
||\begin{align}V &= 6\ \text {V} &I &= 250\ \text {mA} = 0.250\ \text{A} \\ R &= ? \end{align}||
||\begin{align}V = R \times I \quad \Rightarrow \quad R &= \dfrac{V}{I} \\ R &= \dfrac {6\ \text {V}}{0.250\ \text {A}} \\ &= 24\ \Omega \end{align}||
The resistance of the filament is |24\ \Omega.|
What is the current intensity flowing through a |120\ \Omega| resistor subjected to a potential difference of |9\ \text{V}|?
||\begin{align}V &= 9\ \text {V} &R &= 120\ \Omega \\ I &= \text {?} \end{align}||
||\begin{align}V = R \times I \quad \Rightarrow \quad I &= \dfrac{V}{R} \\ I &= \dfrac {9\ \text {V}}{120\ \Omega} \\ &=0.075\ \text {A} \end{align}||
The current intensity flowing through the resistor is |0.075\ \text{A}.|
What is the potential difference at the terminals of a resistance |0.14\ \Omega| wire with a |5\ \text{A}| current flowing through it
||\begin{align}R &= 0.14\ \Omega &I &= 5\ \text {A} \\ V &= \text {?} & \end{align}||
||\begin{align}V = R \times I \quad \Rightarrow \quad V &={0.14\ \Omega}\times {5\ \text {A}} \\ &= 0.7\ \text {V} \end{align}||
The potential difference at the terminals of this wire is |0.7\ \text {V}.|
Pour valider ta compréhension à propos de l'électricité de façon interactive, consulte la MiniRécup suivante :
Pour valider ta compréhension à propos des calculs dans les circuits électriques de façon interactive, consulte la MiniRécup suivante :