Question

Plot 1 in Fig gives the charge q that can be stored on capacitor `C_(1)` versus electric potential V set up across it. Plots 2 and 3 are simillar plots for capacitor `C_(2) and C_(3)` respectively. The three capacitors are connected to 6.0 V battery as shown here. Calculate charge stored in capacitor `C_(2)`.

Answer 1

As is clear from Fig.
`C_(1) = (q_(1))/(V) = (12)/(2) = 6muF`
`C_(2) = (q_(2))/(V) = (8)/(2) = 4 muF`
`C_(3) = (q_(2))/(V) = (4)/(2) = 2muF`
As `C_(2) and C_(3)` are in parallel, therefore
`C_(p) = C_(2) + C_(3) = 4 + 2 = 6muF`
The combination is in seires with `C_(1)`
`:. (1)/(C_(123)) = (1)/(C_(1)) + (1)/(C_(p)) = (1)/(6) + (1)/(6) = (1)/(3)`
`:. C_(123) = 3muF`
Charge on capacitor `C_(1) is q_(1) = 3xx6 = 18 muC`
Voltage across `C_(1) = (q_(1))/(C_(1)) = (18)/(6) = 3 vol t`.
Voltage across `C_(2) = 6-3 = 3 vol t`
`:.` Charge on `C_(2) = 4xx3 = 12 muC`