Two capacitors are connected to a battery. The battery voltage is V = 60 V, and the capacitances are C1 = 2.00 μF and C2 = 4.00 μF. Determine the total energy stored by the two capacitors when they are wired (a) in parallel and (b) in series.

In parallel, C adds,

energy=1/2 CV^2 = 1/2 (6micro)V^2

in series, Ctotal= 8/6=1.333micro
energy= 1/2 (1.33)V^2

where Ctotal is the equivalent capacitor.

To determine the total energy stored by the capacitors, we need to use the formulas for energy stored in a capacitor:

(a) When the capacitors are wired in parallel, the total capacitance (C_parallel) is the sum of the individual capacitances (C1 and C2) since the charge is distributed across both capacitors.

The formula for energy stored in a capacitor is given by:
E = 1/2 * C * V^2

Substituting the given values:
C_parallel = C1 + C2 = 2.00 μF + 4.00 μF = 6.00 μF
V = 60 V

Plugging these values into the formula:
E_parallel = 1/2 * C_parallel * V^2
= 1/2 * 6.00 μF * (60 V)^2
= 180 μJ

So, the total energy stored by the two capacitors when they are wired in parallel is 180 μJ.

(b) When the capacitors are wired in series, the total capacitance (C_series) is given by the reciprocal of the sum of the reciprocals of the individual capacitances:

1/C_series = 1/C1 + 1/C2

Substituting the given values:
1/C_series = 1/2.00 μF + 1/4.00 μF
= 1/2.00 μF + 2/8.00 μF
= 4/8.00 μF + 2/8.00 μF
= 6/8.00 μF
= 0.75/μF

C_series = 1/(0.75/μF)
= 1.33 μF (approximately)

Plugging this value into the formula:
E_series = 1/2 * C_series * V^2
= 1/2 * 1.33 μF * (60 V)^2
= 1.98 mJ (approximately)

So, the total energy stored by the two capacitors when they are wired in series is approximately 1.98 mJ.