Answer
a ) 9 μC
b )11.25 μC , 47.25 μC
Step-by-step explanation
When two capacitor 1.25 μF and 5.25 μF are joined in series , total capacitor is calculated below
a ) C (Total ) =
![(1.25*5.25)/(1.25+5.25)](https://img.qammunity.org/2020/formulas/physics/college/abibe73tckgchlg987ul2h4xrghdbm6ify.png)
= 1 μF
Charge in the circuit
=Potential x capacitance
= 1 x 9 = 9 μC. Each of the capacitor will have this charge.
b ) Charge across 1.25 capacitor = 1.25 x 9 = 11.25 μC
Charge across 5.25 capacitor = 5.25 x 9 = 47.25 μC