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184_notes:c_parallel [2018/06/26 14:42] – [Node Rule and Charge in Parallel] curdemma | 184_notes:c_parallel [2021/03/18 03:07] (current) – [Equivalent Capacitance] bartonmo | ||
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Section 19.1 in Matter and Interactions (4th edition) | Section 19.1 in Matter and Interactions (4th edition) | ||
- | [[184_notes: | + | /*[[184_notes: |
- | [[184_notes: | + | [[184_notes: |
===== Capacitors in Parallel ===== | ===== Capacitors in Parallel ===== | ||
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[{{ 184_notes: | [{{ 184_notes: | ||
- | Just as with resistors, when capacitors are in parallel, this means that there are two separate paths that share the same potential difference. Consider a circuit with a battery and two capacitors that are now parallel to the battery (side-by-side with a junction in the wire). Again, with capacitors, we will look at the charge rather than current since it is the charges that collect on the plates. When we use the node rule at the location where the wire splits, we should then see that any charge that comes into the junction should equal the total charge leaving the junction (remember this is just a statement of the conservation of charge). For Node A (and also for Node B), we can see that the **//total charge coming from the battery should be equal to the charge that travels to the first capacitors plus the charge that travels to the second capacitor//**. | + | Just as with resistors, when capacitors are in parallel, this means that there are two separate paths that share the //same potential difference.// Consider a circuit with a battery and two capacitors that are now parallel to the battery (side-by-side with a junction in the wire). Again, with capacitors, we will look at the //charge// rather than current since it is the charges that collect on the plates. When we use the node rule at the location where the wire splits, we should then see that any charge that comes into the junction should equal the total charge leaving the junction (remember this is just a statement of the conservation of charge). For Node A (and also for Node B), we can see that the **total charge coming from the battery should be equal to the charge that travels to the first capacitors plus the charge that travels to the second capacitor**. |
Qbat=QC1+QC2 | Qbat=QC1+QC2 | ||
==== Loop Rule and Voltage in Parallel ==== | ==== Loop Rule and Voltage in Parallel ==== | ||
- | {{184_notes: | + | [{{184_notes: |
For the parallel circuit with capacitors, we again will now have three loops and thus three equations to check. Again start by marking the high and low potential locations in the circuit. For the first loop going clockwise, we see that there is a gain in potential from the battery and then a drop of potential across the capacitor: | For the parallel circuit with capacitors, we again will now have three loops and thus three equations to check. Again start by marking the high and low potential locations in the circuit. For the first loop going clockwise, we see that there is a gain in potential from the battery and then a drop of potential across the capacitor: | ||
+|ΔVbat|−|ΔVC1|=0 | +|ΔVbat|−|ΔVC1|=0 | ||
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|ΔVC1|=|ΔVC2| | |ΔVC1|=|ΔVC2| | ||
- | This tells us that **//capacitors in parallel have an equal potential difference//**. | + | This tells us that **capacitors in parallel have an equal potential difference**. |
|ΔVbat|=|ΔVC1|=|ΔVC2| | |ΔVbat|=|ΔVC1|=|ΔVC2| | ||
==== Equivalent Capacitance ==== | ==== Equivalent Capacitance ==== | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
Finally, we can find the equivalent capacitance for capacitors in parallel. Again, we will do this by comparing the circuit with the two capacitors to one with a single equivalent capacitor, keeping ΔVbat and Qbat the same in each circuit. From the node rule, we found: | Finally, we can find the equivalent capacitance for capacitors in parallel. Again, we will do this by comparing the circuit with the two capacitors to one with a single equivalent capacitor, keeping ΔVbat and Qbat the same in each circuit. From the node rule, we found: | ||
Qbat=QC1+QC2 | Qbat=QC1+QC2 | ||
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Because the potential differences are all equal in parallel, they cancel out, leaving: | Because the potential differences are all equal in parallel, they cancel out, leaving: | ||
Ceq=C1+C2 | Ceq=C1+C2 | ||
- | Thus, **//for capacitors in parallel, the capacitances will add together//**. | + | Thus, **for capacitors in parallel, the capacitances will add together**. |
==== Examples ==== | ==== Examples ==== | ||
[[: | [[: |