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JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1362354546520:dw @experimentX

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1no let me try it

experimentX
 one year ago
Best ResponseYou've already chosen the best response.0dw:1362354809409:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1362354804183:dw

Xavier
 one year ago
Best ResponseYou've already chosen the best response.0An ok way to remember it is that capacitors in series add like resistors in parallel. Other way around too

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1ok since they're in parallel, the voltages are the same. \[V=V_1=V_2\] \[Q=Q_1+Q_2\] \[C=\frac{Q}{V}\] \[C_1=\frac{Q_1}{V_1}=\frac{Q_1}{V}\] \[C_2=\frac{Q_2}{V_2}=\frac{Q_2}{V}\] \[CV=C_1V+C_2V\] \[C\cancel{V}=C_1\cancel{V}+C_2\cancel V\] C=C_1+C_2

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1but they have different capacitances.....hmmm?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1oh I add the capacitances to get C....the total capacitance of the capacitors in parallel \[C=6 \mu F\]

experimentX
 one year ago
Best ResponseYou've already chosen the best response.0yeah yeah ... parallel is like absolute capitalism!!

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1362355380229:dw LOL!!!!!

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1oh ok let's seedw:1362355422603:dw \[C=\frac{Q}{V}\] \[C_1=\frac{Q}{V_1}\] \[C_2=\frac{Q}{V_2}\] \[Q=Q_1=Q_2\] \[V=V_1+V_2\] \[\frac Q C=\frac{Q}{C_1}+\frac{Q}{C_2}\] \[\frac {\cancel Q} C=\frac{\cancel Q}{C_1}+\frac{\cancel Q}{C_2}\] \[\frac 1 C=\frac 1 {C_1}+\frac 1{C_2}\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac 1 C=\frac 1 6+\frac 2 6=\frac 3 6\] \[C=2\]
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