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V(t) Vc −1+ = (1 - e ¹/RC) -(e-t/RC) Ln(1-V(t)) = ¹/RC Vc V(t) Vc C = = C= -t/(R (Ln(1 – V(!))) Vc -t R(Ln(1 – V(t))

V(t) Vc −1+ = (1 - e ¹/RC) -(e-t/RC) Ln(1-V(t)) = ¹/RC Vc V(t) Vc C = = C= -t/(R (Ln(1 – V(!))) Vc -t R(Ln(1 – V(t))

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Use equations from picture to solve with work.

Below is the transcription of the mathematical derivations involving the charging of a capacitor in an RC circuit. The equations detail the relationship between voltage, time, resistance, and capacitance. 1. The first equation describes the voltage across the capacitor at time \( t \): \[ \frac{V(t)}{V_C} = \left(1 - e^{-t/RC}\right) \] 2. Rearranging the equation, we get: \[ -1 + \frac{V(t)}{V_C} = -\left(e^{-t/RC}\right) \] 3. Taking the natural logarithm on both sides, we have: \[ \ln\left(1 - \frac{V(t)}{V_C}\right) = -\frac{t}{RC} \] 4. Solving for the capacitance \( C \), the equation becomes: \[ C = -t / \left(R \ln\left(1 - \frac{V(t)}{V_C}\right)\right) \] 5. Simplifying, the capacitance \( C \) is given by: \[ C = \frac{-t}{R \ln\left(1 - \frac{V(t)}{V_C}\right)} \] These derivations reflect the behavior of a capacitor's voltage over time in an RC circuit. They are crucial for understanding the exponential nature of charging and discharging processes in electronics.
Transcribed Image Text:Below is the transcription of the mathematical derivations involving the charging of a capacitor in an RC circuit. The equations detail the relationship between voltage, time, resistance, and capacitance. 1. The first equation describes the voltage across the capacitor at time \( t \): \[ \frac{V(t)}{V_C} = \left(1 - e^{-t/RC}\right) \] 2. Rearranging the equation, we get: \[ -1 + \frac{V(t)}{V_C} = -\left(e^{-t/RC}\right) \] 3. Taking the natural logarithm on both sides, we have: \[ \ln\left(1 - \frac{V(t)}{V_C}\right) = -\frac{t}{RC} \] 4. Solving for the capacitance \( C \), the equation becomes: \[ C = -t / \left(R \ln\left(1 - \frac{V(t)}{V_C}\right)\right) \] 5. Simplifying, the capacitance \( C \) is given by: \[ C = \frac{-t}{R \ln\left(1 - \frac{V(t)}{V_C}\right)} \] These derivations reflect the behavior of a capacitor's voltage over time in an RC circuit. They are crucial for understanding the exponential nature of charging and discharging processes in electronics.
**Question 5:** If you wanted to insert a delay in a circuit of 1 ms and you had a 1k resistor, what capacitor would you choose, assuming that the Voltage threshold of the circuit you were connecting to is 50% the 5V supply voltage. **Answer:** 1.44 µF
Transcribed Image Text:**Question 5:** If you wanted to insert a delay in a circuit of 1 ms and you had a 1k resistor, what capacitor would you choose, assuming that the Voltage threshold of the circuit you were connecting to is 50% the 5V supply voltage. **Answer:** 1.44 µF
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