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G₁₂=4.00pF -11- C₁ = 6.00pF Ci= 12. OpF C₂ = 4.00pF C₁ = 6.00pF In the circuit shown in the figure each copocitor has a charge of monitude 5.000 on its plate R=20.02 After the switch Sis clased, what will be the current in the cruit at the instant that the capacitors their have lost 60,0% of initial stored enancy 0

G₁₂=4.00pF -11- C₁ = 6.00pF Ci= 12. OpF C₂ = 4.00pF C₁ = 6.00pF In the circuit shown in the figure each copocitor has a charge of monitude 5.000 on its plate R=20.02 After the switch Sis clased, what will be the current in the cruit at the instant that the capacitors their have lost 60,0% of initial stored enancy 0

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### Circuit Diagram Analysis #### Circuit Description: The diagram features a series circuit with three capacitors (C1, C2, and C3) connected with a resistor (R) and a switch (S). #### Components: - **Capacitors:** - \( C_1 = 12.0 \, \text{pF} \) - \( C_2 = 4.00 \, \text{pF} \) - \( C_3 = 6.00 \, \text{pF} \) - **Resistor:** - \( R = 20.0 \, \Omega \) #### Given Conditions: - Each capacitor is initially charged with a charge magnitude of \( 5.00 \, \text{nC} \) on its plates. #### Problem Statement: - After the switch \( S \) is closed, calculate the current in the circuit at the moment when the capacitors have lost 60.0% of their initial stored energy. #### Constants and Formulas: - Charge of an electron/proton: - \( q_e = -1.60 \times 10^{-19} \, \text{C} = -e \) - Permittivity of free space: - \( \varepsilon_0 = 8.85 \times 10^{-12} \, \frac{\text{C}^2}{\text{N} \cdot \text{m}^2} \) ### Explanation of Concepts: This exercise involves understanding the discharging process of capacitors within a closed circuit and calculating the current based on energy dissipation. Calculating this involves understanding the relationship between stored energy in a capacitor and its discharge rate through a resistive load. ### Learning Objective: - Calculate electric current in discharging circuits. - Apply the concept of energy loss in capacitors. This analysis is intended for students learning about electric circuits and provides a practical application of capacitor discharge equations and energy concepts.
Transcribed Image Text:### Circuit Diagram Analysis #### Circuit Description: The diagram features a series circuit with three capacitors (C1, C2, and C3) connected with a resistor (R) and a switch (S). #### Components: - **Capacitors:** - \( C_1 = 12.0 \, \text{pF} \) - \( C_2 = 4.00 \, \text{pF} \) - \( C_3 = 6.00 \, \text{pF} \) - **Resistor:** - \( R = 20.0 \, \Omega \) #### Given Conditions: - Each capacitor is initially charged with a charge magnitude of \( 5.00 \, \text{nC} \) on its plates. #### Problem Statement: - After the switch \( S \) is closed, calculate the current in the circuit at the moment when the capacitors have lost 60.0% of their initial stored energy. #### Constants and Formulas: - Charge of an electron/proton: - \( q_e = -1.60 \times 10^{-19} \, \text{C} = -e \) - Permittivity of free space: - \( \varepsilon_0 = 8.85 \times 10^{-12} \, \frac{\text{C}^2}{\text{N} \cdot \text{m}^2} \) ### Explanation of Concepts: This exercise involves understanding the discharging process of capacitors within a closed circuit and calculating the current based on energy dissipation. Calculating this involves understanding the relationship between stored energy in a capacitor and its discharge rate through a resistive load. ### Learning Objective: - Calculate electric current in discharging circuits. - Apply the concept of energy loss in capacitors. This analysis is intended for students learning about electric circuits and provides a practical application of capacitor discharge equations and energy concepts.
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