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184_notes:examples:week7_wire_dimensions [2017/10/04 00:49] – [Example: Application of Node Rule] tallpaul | 184_notes:examples:week7_wire_dimensions [2017/10/04 13:51] – [Solution] tallpaul | ||
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=====Example: | =====Example: | ||
- | 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 $I_1 = 5 \text{ A}$, and the voltage across the first two meters is $\Delta V_1 = 1 \text{ V}$. | + | 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 $I_1 = 5 \text{ A}$, and the voltage across the first two meters is $\Delta V_1 = 1 \text{ V}$. In all three segments of the wire, determine the magnitude of the electric field inside and the power transmitted. |
- | + | ||
- | {{ 184_notes: | + | |
===Facts=== | ===Facts=== | ||
- | * $I_1=8 \text{ | + | * Segment lengths: |
- | * $I_1$, $I_2$, and $I_3$ are directed as pictured. | + | * Segment diameters: |
+ | * Current: $I_1 = 5 \text{ A}$. | ||
+ | * Voltage: $\Delta V_1 = 1 \text{ V}$, $\Delta V_{battery} = 12 \text{ V}$. | ||
===Lacking=== | ===Lacking=== | ||
- | * All other currents (including their directions). | + | * Power and electric field in all segments |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The current is not changing (circuit is in steady state). | + | * The circuit is in a steady state. |
- | * All current in the circuit arises from other currents in the circuit. | + | * Approximating |
- | * No resistance in the battery (approximating the battery as a mechanical battery) | + | * The wire has a circular cross-section. |
+ | * No outside influence on the circuit. | ||
+ | * The wire is made of the same material throughout. | ||
===Representations=== | ===Representations=== | ||
- | * We represent the situation with diagram | + | * We represent the situation with diagram |
- | * We represent | + | |
+ | {{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
- | Let's start with node $A$. Incoming current | + | Let's start with segment 1. The electric field is constant since the wire is uniform with respect |
- | $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$ | + | The power dissipated through |
- | + | ||
- | We do a similar analysis for node $B$. Incoming current is $I_{A\rightarrow B}$, and outgoing current is $I_3$. Since $I_{A\rightarrow B}=5 \text{ | + | |
- | $$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$ | + | |
- | + | ||
- | For node $C$, incoming current is $I_2$ and $I_3$. There is no outgoing current defined yet! $I_{C\rightarrow D}$ must be outgoing to balance. To satisfy | + | |
- | $$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$ | + | |
- | + | ||
- | Lastly, we look at node $D$. Incoming current | + | |
- | $$I_{D\rightarrow battery} | + | |
- | + | ||
- | Notice that $I_{D\rightarrow battery}=I_1$. This will always be the case for currents going in and out of the battery (approximating a few things that are usually safe to approximate, | + | |
- | {{ 184_notes:6_nodes_with_arrows.png? | + | Now, for segment 2. We can use [[184_notes:r_energy# |