Question

Capacitors of 5.00 µF, 10.0 µF, and 50.0 µF are connected in series across a 12.0-V battery. What is the potential difference across the 10.0-µF capacitor?

Answer 1

parallel capacitors have the same voltage drop, and the fact that series capacitors have the same charge. thus, we will find the charge of the equivalent capacitance for this series distribution and after calculate the voltage of 10-uF from C = Q / V so:

$\frac{1}{C_{eq} } = \frac{1}{5\mu F} + \frac {1}{10 \mu F} + \frac{1}{50\mu F} = 0.32(\mu F)^{-1} \Rightarrow C_{eq} = 3.125\;\mu F$

Now:

$C_{eq} = \frac{Q}{\Delta V} \Rightarrow Q = C_{eq}\Delta V = (3.125\;\mu F)(12\;V) = 37.5\mu C$

this charge is the same for each capacitor ('cause they are in series distribution), therefore:

$10\;\mu F = \frac{37.5\mu C}{\Delta V_{10\mu F} } \Rightarrow \Delta V_{10\mu F} = \frac{37.5\mu C}{10\;\mu F } = \boxed{3.75\;V}$