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Answer :
Final answer:
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.
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