In each capacitance `C_(1)` is `6.0 muF`, and each capacitance `C_(2)` is `4.0 muF`.
.
The charge on `C_(1)` nearest to `a` when `V_(ab)=420 V` is.

Options

A. `840 muC`
B. `560 muC`
C. `600 muC`
D. `320 muC`

Answer

A

Explanation

Reduction of the farthest right leg yields
`C=(1/(6.0 muF)+1/(6.0 muF)+1/(6.0 muF))^(-1)=20.0 muF=C_(1)/3`
It combines in parallel with `C_(2)`,i.e.,
`C=4.0 muF+2.0 muF=6.0muF=C_(1)`
So the next reduction is the same as the second, leaving `3C_(1)'s`, in series.
So `C_(eq)=2.0 muF=C_(1)//3`. For the three capacitors nearest to points `a` and `b`:
`Q_(C1)=C_(eq)V=(2.0xx10^(-6)F)(420 V)=8.4xx10^(-4)C`
`V_(cd)=1/3((420)/(3)V)=46.7 V`.
By symmetry, the total voltage drop over the equivalent capacitance of the part of the circuit from the junctions between`c` and `d` is `(420//3) V` and the equivalent capacitance is that of one of those capacitors, i.e., `1//3` of `420//3 V`.