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JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1362354546520:dw @experimentX
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
no let me try it
 one year ago

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

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1362354804183:dw
 one year ago

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

JenniferSmart1Best ResponseYou've already chosen the best response.1
ok 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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
but they have different capacitances.....hmmm?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
oh I add the capacitances to get C....the total capacitance of the capacitors in parallel \[C=6 \mu F\]
 one year ago

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

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1362355380229:dw LOL!!!!!
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
oh ok let's seedw:1362355422603:dw \[C=\frac{Q}{V}\] \[C_1=\frac{Q}{V_1}\] \[C_2=\frac{Q}{V_2}\] \[Q=Q_1=Q_2\] \[V=V_1+V_2\] \[\frac Q C=\frac{Q}{C_1}+\frac{Q}{C_2}\] \[\frac {\cancel Q} C=\frac{\cancel Q}{C_1}+\frac{\cancel Q}{C_2}\] \[\frac 1 C=\frac 1 {C_1}+\frac 1{C_2}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[\frac 1 C=\frac 1 6+\frac 2 6=\frac 3 6\] \[C=2\]
 one year ago
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