A lab tech builds a circuit as shown in the figure. Find the following. (Assume $C_1 = 26.0 \, \mu F$ and $C_2 = 3.23 \, \mu F$.)
6.00 µF
$C_1 \, \mu F$
$C_2 \, \mu F$
$C_1 \, \mu F$
9.00 V
(a) the equivalent capacitance (in µF)
54.1
X µF
(b) the charge on each capacitor (in µC)
$C_1$ (left)
234
Χ µC
$C_1$ (right)
234
Χ µC
$C_2$
18.9
Χ µC
6.00 µF capacitor 18.9
Χ µC
(c) the potential difference across each capacitor (in V)
$C_1$ (left)
9
X V
$C_1$ (right)
9
X V
$C_2$
5.85
X V
6.00 µF capacitor 3.15
X V
