Question

Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor.

(a) Find the charge and energy stored if the capacitors are connected to the battery in series.

a) Charge: 0.12 C, Energy: 0.54 J
b) Charge: 0.54 C, Energy: 0.12 J
c) Charge: 0.18 C, Energy: 0.81 J
d) Charge: 0.81 C, Energy: 0.18 J
Do the same for a parallel connection.

a) Charge: 0.81 C, Energy: 0.81 J
b) Charge: 0.18 C, Energy: 0.18 J
c) Charge: 0.54 C, Energy: 0.54 J
d) Charge: 0.12 C, Energy: 0.12 J

Answer 1

For series connection, the total capacitance is 1.56 µF, the charge is 14.04 µC, and the energy stored is 63 mJ. For parallel connection, the total capacitance is 9.4 µF, the charge is 84.6 µC, and energy stored is 380.7 mJ.

Step-by-step explanation:

When connecting capacitors in series, the total capacitance (Ctotal) can be found using the formula: 1/Ctotal = 1/C1 + 1/C2. Using C1 = 2.00 µF and C2 = 7.40 µF, we calculate Ctotal = 1.56 µF. The charge (Q) on capacitors in series is the same and given by Q = Ctotal × V, where V is the voltage, which is 9.00 V in this case. Therefore, Q = 1.56 µF × 9.00 V = 0.01404 C or 14.04 µC. The energy (U) stored in a capacitor is given by U = ½ QV = ½ CtotalV2. Plugging in the numbers, we get U = ½ × 1.56 µF × (9.00 V)2 = 0.063 J or 63 mJ.

In the case of a parallel connection, the total capacitance is a sum of individual capacitances: Ctotal = C1 + C2 = 9.4 µF. The charge on each capacitor is given by Qn = Cn × V, resulting in Q = 9.4 µF × 9.00 V = 0.0846 C or 84.6 µC. The energy stored in the parallel connection is U = ½ × CtotalV2 = ½ × 9.4 µF × (9.00 V)2 = 0.3807 J or 380.7 mJ.