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184_notes:examples:week7_charging_capacitor [2017/10/06 00:16] – [Looking at a Capacitor as it's Charging] tallpaul | 184_notes:examples:week7_charging_capacitor [2017/10/06 18:32] – [Solution] tallpaul | ||
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===Lacking=== | ===Lacking=== | ||
- | * Graphs of $I$, $Q$, $\Delta V$$. | + | * Graphs of $I$, $Q$, $\Delta V$. |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
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====Solution==== | ====Solution==== | ||
- | Let's start with node $A$. Incoming current | + | Before the capacitor |
- | $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$ | + | |
- | We do a similar analysis for node $B$. Incoming current is $I_{A\rightarrow B}$, and outgoing current is $I_3$. Since $I_{A\rightarrow B}=5 \text{ A}$ and $I_3=4 \text{ A}$, we need $I_{B\rightarrow D}$ to be outgoing to balance. To satisfy the Node Rule, we set | + | {{ 184_notes: |
- | $$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ A}$$ | + | |
- | For node $C$, incoming | + | Slightly after time $t=0$, current |
- | $$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ A}$$ | + | |
- | Lastly, we look at node $D$. Incoming current is $I_{B\rightarrow D}$ and $I_{C\rightarrow D}$. Since there is no outgoing current defined yet, $I_{D\rightarrow battery}$ must be outgoing to balance. To satisfy the Node Rule, we set | + | {{ 184_notes: |
- | $$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$ | + | |
- | Notice that $I_{D\rightarrow battery}=I_1$. This will always be the case for currents going in and out of the battery (approximating | + | During this time, the current is smaller than at the beginning, but it is also decreasing at a slower rate, since the plates are not charging |
- | {{ 184_notes:6_nodes_with_arrows.png?300 |Circuit with Nodes}} | + | {{ 184_notes:7_cap_time_3.png?300 |Capacitor |
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+ | Based on what we have argued so far, the graph of the current in the circuit should look something like this: | ||
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+ | {{ 184_notes: | ||
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+ | Notice that we also used the charge of the capacitor in our argument for why the current drops off like it does. We basically argue that the charge graph will look like a flipped current graph: rapidly accumulating charge at first, and gradually slowing down in accumulation until equilibrium (" | ||
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+ | {{ 184_notes: | ||
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+ | Lastly, we wish to produce a graph of the potential difference between the plates. We can pull from a previous in-class project in which we found the electric field from a large plate of charge. It is pretty much constant, and depends on the charge of the plate like so: | ||
+ | $$E_{plate} = \frac{Q_{plate}}{2\epsilon_0 A_{plate}}$$ | ||
+ | We also remember that potential difference across a constant electric field is just the electric field times the distance. Since the distance between the plates is unchanging and the electric field simply scales with $Q$, we can expect the voltage graph to have the same shape as the charge graph: | ||
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+ | {{ 184_notes: | ||
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+ | That's it! We can even do one more check to be sure we are happy with our result. Notice the the capacitance of a capacitor can be represented as the charge on a plate (or something else, depending on the shape of the capacitor) divided by the potential difference between the plates (or whatever): $$C = \frac{Q}{\Delta V}$$ | ||
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+ | We know that capacitance is a constant, since it is an intrinsic property of the capacitor' |