Difference between revisions of "Electrical Resistance"
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| − | | style="height:20px; width:300px; text-align:center;" |'''A student [[measure]]s a [[Potential Difference|potential difference]] of 5. | + | | style="height:20px; width:300px; text-align:center;" |'''A student [[measure]]s a [[Potential Difference|potential difference]] of 5.9V across a [[component]] and a [[Electrical Current|current]] of 0.11A. Calculate the resistance correct to two [[Significant Figures|significant figures]].''' |
| style="height:20px; width:300px; text-align:center;" |'''Calculate the resistance of a [[buzzer]] connected in [[Series Circuit|series]] to a 9V [[battery]] with an [[ammeter]] reading of 23mA correct to two [[Significant Figures|significant figures]].''' | | style="height:20px; width:300px; text-align:center;" |'''Calculate the resistance of a [[buzzer]] connected in [[Series Circuit|series]] to a 9V [[battery]] with an [[ammeter]] reading of 23mA correct to two [[Significant Figures|significant figures]].''' | ||
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Revision as of 15:13, 25 February 2019
Contents
Key Stage 3
Meaning
Resistance is a description of how difficult it is to increase the current through a conductor when increasing the potential difference.
About Resistance
- The unit of resistance is the Ohm (Ω).
- Resistance cannot be directly measured. Resistance must be calculated by dividing the Potential Difference by the Current.
- Conductors have a low resistance and insulators have a high resistance.
Equation
Resistance = (Potential Difference)/(Current)
\[R = \frac{V}{I}\] Where: \[R\] = Resistance of a component. \[V\] = The Potential Difference across the component. \[I\] = The current through the component.
Example Calculations
| A student measures a potential difference of 5V across a component and a current of 0.1A. Calculate the resistance. | A bulb has a current of 200mA passing through it and a potential difference of 12V across it. Calculate the resistance of the bulb. | Calculate the resistance of a buzzer connected in series to a 6V battery with an ammeter reading of 10mA. |
|
Potential Difference = 5V Current = 0.1A \(R = \frac{V}{I}\) \(R = \frac{5}{0.1}\) \( R = 50 \Omega\) |
Potential Difference = 12V Current = 200mA = 0.2A \(R = \frac{V}{I}\) \(R = \frac{12}{0.2}\) \( R = 60 \Omega\) |
Potential Difference = 6V Current = 10mA = 0.01A \(R = \frac{V}{I}\) \(R = \frac{6}{0.01}\) \( R = 600 \Omega\) |
Key Stage 4
Meaning
Resistance is the ratio of potential difference to current.
About Resistance
- The unit of resistance is the Ohm (Ω).
- Resistance cannot be directly measured. Resistance must be calculated by dividing the Potential Difference by the Current.
- Conductors have a low resistance and insulators have a high resistance.
Equation
Resistance = (Potential Difference)/(Current)
\[R = \frac{V}{I}\] Where: \[R\] = Resistance of a component. \[V\] = The Potential Difference across the component. \[I\] = The current through the component.
Example Calculations
| A student measures a potential difference of 5.9V across a component and a current of 0.11A. Calculate the resistance correct to two significant figures. | Calculate the resistance of a buzzer connected in series to a 9V battery with an ammeter reading of 23mA correct to two significant figures. |
| 1. State the known quantities in correct units.
Potential Difference = 5.9V Current = 0.11A |
1. State the known quantities in correct units.
Potential Difference = 9V Current = 23mA = 23x10-3A |
| 2. Substitute the numbers into the equation and solve.
\(R = \frac{V}{I}\) \(R = \frac{5.9}{0.11}\) \( R = 53.635 \Omega\) \( R \approx 54\Omega\) |
2. Substitute the numbers into the equation and solve.
\(R = \frac{V}{I}\) \(R = \frac{9}{23\times10^{-3}}\) \( R = 391.304 \Omega\) \( R \approx 390 \Omega\) |