## anonymous 3 years ago Capacitors in series and in parallel

1. anonymous

|dw:1362354546520:dw| @experimentX

2. anonymous

no let me try it

3. experimentX

|dw:1362354809409:dw|

4. anonymous

|dw:1362354804183:dw|

5. anonymous

oh ok

6. anonymous

An ok way to remember it is that capacitors in series add like resistors in parallel. Other way around too

7. anonymous

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

8. anonymous

but they have different capacitances.....hmmm?

9. anonymous

oh I add the capacitances to get C....the total capacitance of the capacitors in parallel $C=6 \mu F$

10. experimentX

yeah yeah ... parallel is like absolute capitalism!!

11. anonymous

|dw:1362355380229:dw| LOL!!!!!

12. anonymous

oh ok let's see|dw:1362355422603:dw| $C=\frac{Q}{V}$ $C_1=\frac{Q}{V_1}$ $C_2=\frac{Q}{V_2}$ $Q=Q_1=Q_2$ $V=V_1+V_2$ $\frac Q C=\frac{Q}{C_1}+\frac{Q}{C_2}$ $\frac {\cancel Q} C=\frac{\cancel Q}{C_1}+\frac{\cancel Q}{C_2}$ $\frac 1 C=\frac 1 {C_1}+\frac 1{C_2}$

13. anonymous

$\frac 1 C=\frac 1 6+\frac 2 6=\frac 3 6$ $C=2$

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