If [tex]C_1 = C_2 = 4.00 \, \mu F[/tex] and [tex]C_4 = 8.00 \, \mu F[/tex], what must the capacitance [tex]C_3[/tex] be for the network to store [tex]2.70 \times 10^{-3} \, J[/tex] of electrical energy?

To find the capacitance C₃ that allows the network to store 2.70x10⁻³ J of electrical energy, we need to calculate the equivalent capacitance of the network. The total capacitance of the network is 10.67 μF, therefore the capacitance C₃ must be 10.67 μF for the network to store the given amount of electrical energy.

Explanation:

To find the capacitance C₃ that allows the network to store 2.70x10⁻³ J of electrical energy, we need to calculate the equivalent capacitance of the network. Since C₁ and C₂ are connected in series, their total capacitance can be found using the formula 1/Ct = 1/C₁ + 1/C₂. Substituting the given values, we get Ct = 2.67 μF.

Next, we combine this equivalent capacitance with C₄, which is connected in parallel. The total capacitance C of the network is given by C = Ct + C₄ = 2.67 μF + 8.00 μF = 10.67 μF.

Therefore, the capacitance C₃ must be 10.67 μF for the network to store 2.70x10⁻³ J of electrical energy.