Question

Question

Answer 1

\[\frac 1 C=\frac 1 C_1 +\frac 1 C_2\]
\[\frac 1 C=\frac{C_2+C_1}{C_1C_2}\]
\[C_{eq}=\frac{C_1C_2}{C_2+C_1}\]
\[C=C_3+C_{eq}\]
\[C=C_3+\frac{C_1C_2}{C_2+C_1}\]

Answer 2

@UnkleRhaukus

Answer 3

@.Sam.

Answer 4

looks right to me \[\checkmark\]

Answer 5

oh cool!

Answer 6

should I solve for Q for each capacitor (for part b)
since I know C and V should be 200 volts for each pair of plates, correct?

Answer 7

i think thats it

Answer 8

ok

Answer 9

why?

Answer 10

due to the fact that some of them are in parallel and some in series?

Answer 11

I think its fine, once u got the capacitance then u can use Q=CV to find charge

Answer 12

Remember each capacitor has different voltages

Answer 13

But u can tell that C3 is 200V and for C1 and C2 is V1+V2=200

Answer 14

does that mean that capacitors in series have to share the voltage (in this case) because they're in this circuit