## JenniferSmart1 2 years ago Capacitors in series and in parallel

1. JenniferSmart1

|dw:1362354546520:dw| @experimentX

2. JenniferSmart1

no let me try it

3. experimentX

|dw:1362354809409:dw|

4. JenniferSmart1

|dw:1362354804183:dw|

5. JenniferSmart1

oh ok

6. Xavier

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

7. JenniferSmart1

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

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

9. JenniferSmart1

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. JenniferSmart1

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

12. JenniferSmart1

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

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