I was provided with the following problem.
So I first calculated the total capacitance for (i), which was 4.5+1.5=6.0μF
Now part (ii) is the question i'm struggling at. I know that the charge on the 4.5μF capacitor is 6.3×4.5=28.35μF
So why is the p.d. across both capacitors equal to
V=QC=28.35×10−6(4.5+1.5)×10−6=4.7V
I understand that the charge has to be the same on both plates but why do you add the capacitance values together? Wouldn't the different values of capacitances (C) mean that V=QC are different across each capacitor?
No comments:
Post a Comment