[{{ 184_notes:Week8_13.png?350|Node rule representation for a circuit with capacitors in parallel}}]
[{{ 184_notes:Week8_13.png?350|Node rule representation for a circuit with capacitors in parallel}}]
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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**.
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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
Line 27:
Line 27:
|ΔVC1|=|ΔVC2|
|ΔVC1|=|ΔVC2|
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This tells us that **//capacitors in parallel have an equal potential difference//**.
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This tells us that **capacitors in parallel have an equal potential difference**.
|ΔVbat|=|ΔVC1|=|ΔVC2|
|ΔVbat|=|ΔVC1|=|ΔVC2|
Line 42:
Line 42:
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
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Thus, **//for capacitors in parallel, the capacitances will add together//**.
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Thus, **for capacitors in parallel, the capacitances will add together**.