Question

I need help solving my homework problem , can somebody help me please . This class is physics two . My question is “ where the potential difference across ab is 49.5 V “ ?

Answer 1

Step-by-step explanation:

a)
`Q_(Total) = C_(eq)V = (3.5\:\mu \text{F})(49.5V) = 1.73×10^(-4)\:\text{C}`

b) The individual capacitors in series circuit carry the same amount of charge as the
`Q_(Total)` so

`Q_(10) = Q_(Total) = 1.73×10^(-4)\:\text{C}`

c) Likewise,
`C_9` will carry the same amount of charge as
`C_(10)`