Difference between revisions of "Electrical Resistance"

Revision as of 14:43, 25 February 2019

Resistance is a description of how difficult it is to increase the current through a conductor when increasing the potential difference.

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.

\[Resistance = \frac{Potential Difference}{Current}\] \[R = \frac{V}{I}\] Where:

R = Resistance

V = Potential Difference

I = Current

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 \(R = \frac{V}{I}\) \(R = \frac{6}{0.01}\) \( R = 600 \Omega\)

@@ Line 28: / Line 28: @@
 Current = 0.1[[A]]
-<math>R = \tfrac{V}{I}</math>
+<math>R = \frac{V}{I}</math>
-<math>R = \tfrac{5}{0.1}</math>
+<math>R = \frac{5}{0.1}</math>
 <math> R = 50 \Omega</math>
@@ Line 38: / Line 38: @@
 Current = 200[[mA]] = 0.2[[A]]
-<math>R = \tfrac{V}{I}</math>
+<math>R = \frac{V}{I}</math>
-<math>R = \tfrac{12}{0.2}</math>
+<math>R = \frac{12}{0.2}</math>
 <math> R = 60 \Omega</math>
@@ Line 48: / Line 48: @@
 Current = 10[[mA]] = 0.01[[A]]
-<math>R = \tfrac{V}{I}</math>
+<math>R = \frac{V}{I}</math>
-<math>R = \tfrac{6}{0.01}</math>
+<math>R = \frac{6}{0.01}</math>
 <math> R = 600 \Omega</math>
 |}