## JenniferSmart1 Group Title Capacitors in series and in parallel one year ago one year ago

1. JenniferSmart1 Group Title

|dw:1362354546520:dw| @experimentX

2. JenniferSmart1 Group Title

no let me try it

3. experimentX Group Title

|dw:1362354809409:dw|

4. JenniferSmart1 Group Title

|dw:1362354804183:dw|

5. JenniferSmart1 Group Title

oh ok

6. Xavier Group Title

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

7. JenniferSmart1 Group Title

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. JenniferSmart1 Group Title

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

9. JenniferSmart1 Group Title

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

10. experimentX Group Title

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

11. JenniferSmart1 Group Title

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

12. JenniferSmart1 Group Title

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. JenniferSmart1 Group Title

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