Difference between revisions of "Electrical Resistance"
| Line 15: | Line 15: | ||
: V = Potential Difference | : V = Potential Difference | ||
: I = Current | : I = Current | ||
| + | |||
| + | Calculate the solutions to the worded questions: | ||
| + | a. | ||
| + | b. A bulb has a current of 200mA passing through it and a potential difference of 12V across it. Calculate the resistance of the bulb. | ||
| + | c. State the resistance of a buzzer connected in series to a 6V battery with an ammeter reading of 10mA. | ||
| + | |||
| + | |||
| + | ===Example Calculations=== | ||
| + | {| class="wikitable" | ||
| + | |- | ||
| + | | style="height:20px; width:200px; text-align:center;" |'''A student [[measure]]s a [[Potential Difference|potential difference]] of 5[[V]] across a [[component]] and a [[Electrical Current|current]] of 0.1[[A]]. Calculate the resistance.''' | ||
| + | | style="height:20px; width:200px; text-align:center;" |'''A [[Electrical Bulb|bulb]] has a [[Electrical Current|current]] of 200[[mA]] passing through it and a [[Potential Difference|potential difference]] of 12[[V]] across it. Calculate the resistance of the [[Electrical Bulb|bulb]].''' | ||
| + | | style="height:20px; width:200px; text-align:center;" |'''Calculate the resistance of a [[buzzer]] connected in [[Series Circuit|series]] to a 6[[V]] [[battery]] with an [[ammeter]] reading of 10[[mA]].''' | ||
| + | |- | ||
| + | | style="height:20px; width:200px; text-align:left;" | | ||
| + | Force = 40N | ||
| + | |||
| + | Distance moved in the direction of the force = 1.2m | ||
| + | |||
| + | <math> W = Fd</math> | ||
| + | |||
| + | <math> W = 40 \times 1.2</math> | ||
| + | |||
| + | <math> W = 48J</math> | ||
| + | | style="height:20px; width:200px; text-align:left;" | | ||
| + | Force = 10,000N | ||
| + | |||
| + | Distance moved in the direction of the force = 15m | ||
| + | |||
| + | <math> W = Fd</math> | ||
| + | |||
| + | <math> W = 10,000 \times 15</math> | ||
| + | |||
| + | <math> W = 15,000J</math> | ||
| + | | style="height:20px; width:200px; text-align:left;" | | ||
| + | Force = 40N | ||
| + | |||
| + | Distance moved in the direction of the force = 0m | ||
| + | |||
| + | <math> W = Fd</math> | ||
| + | |||
| + | <math> W = 20 \times 0</math> | ||
| + | |||
| + | <math> W = 0J</math> | ||
| + | |||
| + | No '''work''' has been done because the movement is not in the direction of the [[force]]. The [[weight]] acts downwards but the movement was horizontal. | ||
| + | |} | ||
Revision as of 08:23, 26 October 2018
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
\[Resistance = \tfrac{Potential Difference}{Current}\] \[R = \tfrac{V}{I}\] Where:
- R = Resistance
- V = Potential Difference
- I = Current
Calculate the solutions to the worded questions: a. b. A bulb has a current of 200mA passing through it and a potential difference of 12V across it. Calculate the resistance of the bulb. c. State the resistance of a buzzer connected in series to a 6V battery with an ammeter reading of 10mA.
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. |
|
Force = 40N Distance moved in the direction of the force = 1.2m \( W = Fd\) \( W = 40 \times 1.2\) \( W = 48J\) |
Force = 10,000N Distance moved in the direction of the force = 15m \( W = Fd\) \( W = 10,000 \times 15\) \( W = 15,000J\) |
Force = 40N Distance moved in the direction of the force = 0m \( W = Fd\) \( W = 20 \times 0\) \( W = 0J\) No work has been done because the movement is not in the direction of the force. The weight acts downwards but the movement was horizontal. |