184_notes:examples:week7_wire_dimensions

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184_notes:examples:week7_wire_dimensions [2018/06/19 14:18] – [Example: Changing the Dimensions of a Wire] curdemma184_notes:examples:week7_wire_dimensions [2021/06/14 23:40] (current) – schram45
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 +[[184_notes:r_energy|Return to Energy in Circuits]]
 +
 =====Example: Changing the Dimensions of a Wire===== =====Example: Changing the Dimensions of a Wire=====
 Suppose you have a simple circuit whose wire changes in thickness. The wire is 8 meters long. The first 2 meters of the wire are 3 mm thick. The next 2 meters are 1 mm thick. The last 4 meters are 3 mm thick. The wire is connected to a 12-Volt battery and current is allowed to flow. You use an ammeter and a voltmeter to find that the current through the first 2 meters of wire is I1=5 A, and the voltage across the first two meters is ΔV1=1 V. In all three segments of the wire, determine the magnitude of the electric field inside and the power transmitted. Suppose you have a simple circuit whose wire changes in thickness. The wire is 8 meters long. The first 2 meters of the wire are 3 mm thick. The next 2 meters are 1 mm thick. The last 4 meters are 3 mm thick. The wire is connected to a 12-Volt battery and current is allowed to flow. You use an ammeter and a voltmeter to find that the current through the first 2 meters of wire is I1=5 A, and the voltage across the first two meters is ΔV1=1 V. In all three segments of the wire, determine the magnitude of the electric field inside and the power transmitted.
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 ===Approximations & Assumptions=== ===Approximations & Assumptions===
-  * The circuit is in a steady state - which means the current should be the same in all three sections. +  * The circuit is in a steady state: This allows the current should be the same in all three sections. 
-  * Approximating the battery as a mechanical battery. +  * Approximating the battery as a mechanical battery: Batteries normally dont keep their energy forever and slowly die over time. Using a mechanical battery means our battery produces a steady source of energy in this problem. 
-  * The wire has a circular cross-section. +  * No outside influence on the circuit: This simplifies the model and isolates our circuit from any outside sources of charge or energy that could effect our calculations.
-  * No outside influence on the circuit. +
-  * The wire is made of the same material throughout.+
  
 ===Representations=== ===Representations===
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 Now, for segment 2. We can use [[184_notes:r_energy#Conservation_of_Charge_in_Circuits|what we know]] about charge in steady state circuits to determine the electric field (notice we divide diameter by 2 in order to get the radius of the circular cross-section): E2=A1A2E1=π(d1/2)2π(d2/2)2E1=9E1=4.5 V/m
 Now, for segment 2. We can use [[184_notes:r_energy#Conservation_of_Charge_in_Circuits|what we know]] about charge in steady state circuits to determine the electric field (notice we divide diameter by 2 in order to get the radius of the circular cross-section): E2=A1A2E1=π(d1/2)2π(d2/2)2E1=9E1=4.5 V/m
 +
 +<WRAP TIP>
 +===Assumptions====
 +In order to do this calculation there are two important assumptions that must be made
 +  * The wires have a circular cross section: This allows us to use the formula for the area of a circle to come up with the correct proportion.
 +  * The wires are made of the same material throughout: There are two terms in the electron current equation that are material propeties and these will cancel out for each segment of wire if they are made of the same material. This allows the electric field to only vary with cross sectional area.
 +</WRAP>
 +
 A simple application of the [[184_notes:current#Current_in_Different_Parts_of_the_Wire|Current Node Rule]] tells us that I2=I1. The voltage is easily found from the constant electric field: ΔV2=E2L2=9 V. The power dissipated through the segment is then P2=I2ΔV2=45 W
 A simple application of the [[184_notes:current#Current_in_Different_Parts_of_the_Wire|Current Node Rule]] tells us that I2=I1. The voltage is easily found from the constant electric field: ΔV2=E2L2=9 V. The power dissipated through the segment is then P2=I2ΔV2=45 W
  
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