Resistivity - Revision history

Bmduffy5: /* A Mathematical Model */

2016-11-28T00:11:19Z

A Mathematical Model

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	[[File:Ohmic_Resistor1.jpg]]		[[File:Ohmic_Resistor1.jpg]]

	[[File:~~Res2~~.jpg]]		[[File:Resistors1.jpg]]

	The definition provided above is specific to ohmic resistors, as stated. These resistors have a uniform cross-section, where current flows uniformly through them. Instead, a more general definition starts with the idea that an electric field inside a specific material is responsible for the electric current flowing within it. Because most of the resistors we will deal with are uniform, a simple ratio of field to current can be used in regards to resistivity. Thus, the electrical resistivity, or "p" can be defined as the ratio of the electric field to the density of the current it creates:		The definition provided above is specific to ohmic resistors, as stated. These resistors have a uniform cross-section, where current flows uniformly through them. Instead, a more general definition starts with the idea that an electric field inside a specific material is responsible for the electric current flowing within it. Because most of the resistors we will deal with are uniform, a simple ratio of field to current can be used in regards to resistivity. Thus, the electrical resistivity, or "p" can be defined as the ratio of the electric field to the density of the current it creates:

Bmduffy5: /* A Mathematical Model */

2016-11-28T00:09:36Z

A Mathematical Model

← Older revision		Revision as of 20:09, 27 November 2016
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	In a circuit the Electrical Resistance is then used to calculate the current in a circuit using Ohm's Law and the following equation <math>I = \frac{\|\Delta V\|}{R} </math> where '''V''' is the voltage, '''I''' is the current, and '''R''' is the resistance. In this equations voltage and resistance are independent variables, whereas the Current is the dependent variable. This law, while useful, only works for ohmic resistors.		In a circuit the Electrical Resistance is then used to calculate the current in a circuit using Ohm's Law and the following equation <math>I = \frac{\|\Delta V\|}{R} </math> where '''V''' is the voltage, '''I''' is the current, and '''R''' is the resistance. In this equations voltage and resistance are independent variables, whereas the Current is the dependent variable. This law, while useful, only works for ohmic resistors.

	[[File:~~Ohmic_Resistor~~.jpg]]		[[File:Ohmic_Resistor1.jpg]]

	[[File:Res2.jpg]]		[[File:Res2.jpg]]

Bmduffy5 at 00:07, 28 November 2016

2016-11-28T00:07:13Z

← Older revision		Revision as of 20:07, 27 November 2016
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	In a circuit the Electrical Resistance is then used to calculate the current in a circuit using Ohm's Law and the following equation <math>I = \frac{\|\Delta V\|}{R} </math> where '''V''' is the voltage, '''I''' is the current, and '''R''' is the resistance. In this equations voltage and resistance are independent variables, whereas the Current is the dependent variable. This law, while useful, only works for ohmic resistors.		In a circuit the Electrical Resistance is then used to calculate the current in a circuit using Ohm's Law and the following equation <math>I = \frac{\|\Delta V\|}{R} </math> where '''V''' is the voltage, '''I''' is the current, and '''R''' is the resistance. In this equations voltage and resistance are independent variables, whereas the Current is the dependent variable. This law, while useful, only works for ohmic resistors.

	[[File:~~Res1~~.jpg]]		[[File:Ohmic_Resistor.jpg]]

	[[File:Res2.jpg]]		[[File:Res2.jpg]]

Bmduffy5: /* Difficult */

2016-11-27T23:33:47Z

Difficult

← Older revision		Revision as of 19:33, 27 November 2016
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	====Problem 3====		====Problem 3====

	A battery and resistor circuit is connected to a very sensitive ohmmeter and is taken outside and left in the sun on a very hot day. What, if anything, will happen to its reading after being outside for a few minutes and why? Assume the battery is unaffected.		A battery and resistor circuit is connected to a very sensitive ohmmeter and the two are then taken outside and left in the sun on a very hot day. What, if anything, will happen to its reading after being outside for a few minutes and why? Assume the battery is unaffected.

	====Solution====		====Solution====

	The current would be less ~~that~~ it was inside. Since the circuit ~~was taken~~ outside the resistor ~~would~~ heat up due to the ~~sun~~. ~~This would in turn cause its~~ resistance ~~to go~~ up~~. When~~ the ~~resistance goes up~~ and the ~~voltage~~ of the ~~battery stays~~ the ~~same.~~ due to ~~Ohms Law~~ the current ~~must go down, resulting in a lower reading~~.		The current reading would be less than what it was while inside. Since the resistor circuit is outside, and hotter than usual the resistor will heat up. As discussed before, resistivity of a material goes up when it is hotter than normal, due to the increase motion of nuclei. The resistance goes up, but because the battery remains unaffected and holds the same amount of charge, the value of the current must go down (due to the inverse relationship of resistance and current). This is all based off of the very simple Ohms Law; <math>I = \frac{V}{R} </math>

	==Scope==		==Scope==

Bmduffy5: /* Scope */

2016-11-27T23:28:14Z

Scope

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	==Scope==		==Scope==
	Resistivity is very important to electrical ~~engineers~~ and ~~others who work with circuits because it~~ is important to ~~understand~~ how ~~a circuit is going to work when~~ outside of laboratory conditions~~. Users cannot expect their circuits~~ to ~~work the same in both nominal~~ and extreme temperature conditions.
			Resistivity is a very important topic for all types of engineers, especially electrical and mechanical. It is important to know how circuits operate outside of laboratory conditions, and one thing that changes a lot in the real world is resistivity due to normal and extreme temperature conditions. Also, when building things like phones and other electronics, the total cost is always important. Though gold and silver may have very low resistivity, designers have to look at other metals that offer similar attributes but for lower costs. So overall, it is an important fundamental concept for engineers to understand the idea of resistance, but also play a large role in large scale manufacturing of different devices.

	== See also ==		== See also ==

Bmduffy5: /* Solution */

2016-11-27T23:21:43Z

Solution

← Older revision		Revision as of 19:21, 27 November 2016
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	Using the equation <math>I = \frac{dV}{R} </math> we can substitute is 1 for I and 3 for dV leaving us with the equation <math>1 = \frac{3}{R} </math>. Solving for R we come to the conclusion that the resistance must be 3 ohms.		Using the equation <math>I = \frac{dV}{R} </math> we can substitute is 1 for I and 3 for dV leaving us with the equation <math>1 = \frac{3}{R} </math>. Solving for R we come to the conclusion that the resistance must be 3 ohms.

	To calculate resistivity, we put 7.06E-4m^2 for cross-sectional area, 0.06m for length and 3 ohms for resistance into our equation, getting <math>3 = \frac{\rho 0.06}{7.06E-4} </math>. When you solve for resitivity, you get the answer of ~~254~~.65 ohms/meter		To calculate resistivity, we put 7.06E-4m^2 for cross-sectional area, 0.06m for length and 3 ohms for resistance into our equation, getting <math>3 = \frac{\rho 0.06}{7.06E-4} </math>. When you solve for resitivity, you get the answer of 0.0353 ohms/meter

	==='''Middling'''===		==='''Middling'''===

Bmduffy5: /* Simple */

2016-11-27T23:20:17Z

Simple

← Older revision		Revision as of 19:20, 27 November 2016
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	An unknown ohmic resistor is attached to a 3V battery and the current is measured at 1 amp. Calculate the resistance of the unknown resistor.		An unknown ohmic resistor is attached to a 3V battery and the current is measured at 1 amp. Calculate the resistance of the unknown resistor.

	Also, if this resistor has a diameter of 0.03m, and length of 0.06m, what is its resistivity?		Also, theoretically, if this resistor has a diameter of 0.03m, and length of 0.06m, what is its resistivity?

	====Solution====		====Solution====

	Using the equation <math>I = \frac{dV}{R} </math> we can substitute is 1 for I and 3 for dV leaving us with the equation <math>1 = \frac{3}{R} </math>. Solving for R we come to the conclusion that the resistance must be 3 ohms.		Using the equation <math>I = \frac{dV}{R} </math> we can substitute is 1 for I and 3 for dV leaving us with the equation <math>1 = \frac{3}{R} </math>. Solving for R we come to the conclusion that the resistance must be 3 ohms.

			To calculate resistivity, we put 7.06E-4m^2 for cross-sectional area, 0.06m for length and 3 ohms for resistance into our equation, getting <math>3 = \frac{\rho 0.06}{7.06E-4} </math>. When you solve for resitivity, you get the answer of 254.65 ohms/meter

	==='''Middling'''===		==='''Middling'''===

Bmduffy5: /* Solution */

2016-11-27T23:14:33Z

Solution

← Older revision		Revision as of 19:14, 27 November 2016
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	====Solution====		====Solution====

	Using the equation <math>I = \frac{dV}{R} </math> we can substitute is 1 for I and 3 for dV leaving us with the equation 1=3/R. Solving for R we come to the ~~answer~~ that the it must be a 3 ~~ohm resistor~~.		Using the equation <math>I = \frac{dV}{R} </math> we can substitute is 1 for I and 3 for dV leaving us with the equation <math>1 = \frac{3}{R} </math>. Solving for R we come to the conclusion that the resistance must be 3 ohms.

	==='''Middling'''===		==='''Middling'''===

Bmduffy5: /* Solution */

2016-11-27T23:13:20Z

Solution

← Older revision		Revision as of 19:13, 27 November 2016
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	====Solution====		====Solution====

	Using the equation <math>I = \frac{\dV}{R} </math> we can substitute is 1 for I and 3 for dV leaving us with the equation 1=3/R. Solving for R we come to the answer that the it must be a 3 ohm resistor.		Using the equation <math>I = \frac{dV}{R} </math> we can substitute is 1 for I and 3 for dV leaving us with the equation 1=3/R. Solving for R we come to the answer that the it must be a 3 ohm resistor.

	==='''Middling'''===		==='''Middling'''===

Bmduffy5: /* Solution */

2016-11-27T23:12:59Z

Solution

← Older revision		Revision as of 19:12, 27 November 2016
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	====Solution====		====Solution====

	Using the equation <math> I=\|dV\|/R <math> we can substitute is 1 for I and 3 for dV leaving us with the equation 1=3/R. Solving for R we come to the answer that the it must be a 3 ohm resistor.		Using the equation <math>I = \frac{\dV}{R} </math> we can substitute is 1 for I and 3 for dV leaving us with the equation 1=3/R. Solving for R we come to the answer that the it must be a 3 ohm resistor.

	==='''Middling'''===		==='''Middling'''===