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The capacitor stores 79.2 mJ of energy after a Voat = 10.0 V battery has been connected to the circuit for a long time. R. Then, the capacitor discharges half of this stored energy in R, bat exactly 1.28 s when the battery is removed and replaced by a RI = 1.10 x 10³ Q load. C, Determine the value of the capacitance C¡ in microfarads and resistance Rj in ohms.

The capacitor stores 79.2 mJ of energy after a Voat = 10.0 V battery has been connected to the circuit for a long time. R. Then, the capacitor discharges half of this stored energy in R, bat exactly 1.28 s when the battery is removed and replaced by a RI = 1.10 x 10³ Q load. C, Determine the value of the capacitance C¡ in microfarads and resistance Rj in ohms.

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A small circuit consists of a resistor in series with a capacitor, as shown.
 

The capacitor stores 79.2 mJ of energy after a ?bat=10.0 V battery has been connected to the circuit for a long time. Then, the capacitor discharges half of this stored energy in exactly1.28 s when the battery is removed and replaced by a ??=1.10×10^3 Ω load.

Determine the value of the capacitance C1 in microfarads and resistance R1 in ohms.

 

Please solve for both C1 and R1, thanks!

**Series RC Circuit Analysis: Capacitor Discharge** A small circuit consists of a resistor in series with a capacitor, as shown in the diagram below. ### Circuit Description - The diagram shows a resistor \( R_1 \) in series with a capacitor \( C_1 \), connected to a voltage source \( V_{\text{bat}} \). - The circuit includes a switch that, when closed, allows the capacitor to charge. - After removing the voltage source, the capacitor discharges through a load resistor \( R_L \). ### Problem Statement The capacitor stores 79.2 mJ (millijoules) of energy after being connected to a \( V_{\text{bat}} \) = 10.0 V battery for a long time. The stored energy is half-discharged in 1.28 seconds when the battery is replaced by a \( R_L \) = 1.10 × 10³ Ω load. #### Goal Determine the values of the capacitance \( C_1 \) in microfarads (µF) and the resistance \( R_1 \) in ohms (Ω). ### Key Formulas and Concepts - The energy stored in the capacitor, \( E \), is given by: \[ E = \frac{1}{2} C_1 V_{\text{bat}}^2 \] - The time constant \( \tau \) of an RC circuit is: \[ \tau = R_{\text{eq}} C_1 \] where \( R_{\text{eq}} = R_1 + R_L \) (equivalent resistance during discharge). - The voltage across a discharging capacitor as a function of time is: \[ V(t) = V_0 e^{-t/\tau} \] where \( V_0 \) is the initial voltage across the capacitor. ### Given Data - \( V_{\text{bat}} \) = 10.0 V - Stored Energy \( E \) = 79.2 mJ = 79.2 × 10⁻³ J - Time to discharge half the energy \( t \) = 1.28 s - Load Resistor \( R_L \) = 1.10 × 10³ Ω ### Calculation Steps 1. Calculate the capacitance \( C_1 \): \
Transcribed Image Text:**Series RC Circuit Analysis: Capacitor Discharge** A small circuit consists of a resistor in series with a capacitor, as shown in the diagram below. ### Circuit Description - The diagram shows a resistor \( R_1 \) in series with a capacitor \( C_1 \), connected to a voltage source \( V_{\text{bat}} \). - The circuit includes a switch that, when closed, allows the capacitor to charge. - After removing the voltage source, the capacitor discharges through a load resistor \( R_L \). ### Problem Statement The capacitor stores 79.2 mJ (millijoules) of energy after being connected to a \( V_{\text{bat}} \) = 10.0 V battery for a long time. The stored energy is half-discharged in 1.28 seconds when the battery is replaced by a \( R_L \) = 1.10 × 10³ Ω load. #### Goal Determine the values of the capacitance \( C_1 \) in microfarads (µF) and the resistance \( R_1 \) in ohms (Ω). ### Key Formulas and Concepts - The energy stored in the capacitor, \( E \), is given by: \[ E = \frac{1}{2} C_1 V_{\text{bat}}^2 \] - The time constant \( \tau \) of an RC circuit is: \[ \tau = R_{\text{eq}} C_1 \] where \( R_{\text{eq}} = R_1 + R_L \) (equivalent resistance during discharge). - The voltage across a discharging capacitor as a function of time is: \[ V(t) = V_0 e^{-t/\tau} \] where \( V_0 \) is the initial voltage across the capacitor. ### Given Data - \( V_{\text{bat}} \) = 10.0 V - Stored Energy \( E \) = 79.2 mJ = 79.2 × 10⁻³ J - Time to discharge half the energy \( t \) = 1.28 s - Load Resistor \( R_L \) = 1.10 × 10³ Ω ### Calculation Steps 1. Calculate the capacitance \( C_1 \): \
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