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Solving for R using the Discharging curve equation: V(t) = 1.83V Vc = 4.96V C =4.7uF T = 50s V(t) = Vc(e ¹/RC) V(t) Vc (e t/RC) Answer: About 10 MQ Vc -t/RC R= = R=-t/(C(ln(V(t))) You end up with a negative term in the denominator to cancel out the negative term on top. Question 1: Calculate the value of R if the other values are as follows:(show your steps.) -t C(In(V(t)) Vc Question 2: Calculate T (tau) which is the resistor times the capacitor (TRC). This is one time constant for these values.

Solving for R using the Discharging curve equation: V(t) = 1.83V Vc = 4.96V C =4.7uF T = 50s V(t) = Vc(e ¹/RC) V(t) Vc (e t/RC) Answer: About 10 MQ Vc -t/RC R= = R=-t/(C(ln(V(t))) You end up with a negative term in the denominator to cancel out the negative term on top. Question 1: Calculate the value of R if the other values are as follows:(show your steps.) -t C(In(V(t)) Vc Question 2: Calculate T (tau) which is the resistor times the capacitor (TRC). This is one time constant for these values.

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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Please solve and show work for questions 1 and 2 using formulas provided in top half of picture, thank you dearly.

**Solving for R using the Discharging Curve Equation:** \[ V(t) = V_c(e^{-t/RC}) \] \[ \frac{V(t)}{V_c} = (e^{-t/RC}) \] \[ \ln\left(\frac{V(t)}{V_c}\right) = \frac{-t}{RC} \] \[ R = -t/(C\ln(V(t)/V_c)) \] \[ R = \frac{-t}{C \ln\left(\frac{V(t)}{V_c}\right)} \] *You end up with a negative term in the denominator to cancel out the negative term on top.* **Question 1:** Calculate the value of R if the other values are as follows: (show your steps.) - \( V(t) = 1.83V \) - \( V_c = 4.96V \) - \( C = 4.7\mu F \) - \( T = 50s \) **Answer:** About 10 MΩ **Question 2:** Calculate \( \tau \) (tau) which is the resistor times the capacitor (\( \tau = RC \)). This is one time constant for these values.
Transcribed Image Text:**Solving for R using the Discharging Curve Equation:** \[ V(t) = V_c(e^{-t/RC}) \] \[ \frac{V(t)}{V_c} = (e^{-t/RC}) \] \[ \ln\left(\frac{V(t)}{V_c}\right) = \frac{-t}{RC} \] \[ R = -t/(C\ln(V(t)/V_c)) \] \[ R = \frac{-t}{C \ln\left(\frac{V(t)}{V_c}\right)} \] *You end up with a negative term in the denominator to cancel out the negative term on top.* **Question 1:** Calculate the value of R if the other values are as follows: (show your steps.) - \( V(t) = 1.83V \) - \( V_c = 4.96V \) - \( C = 4.7\mu F \) - \( T = 50s \) **Answer:** About 10 MΩ **Question 2:** Calculate \( \tau \) (tau) which is the resistor times the capacitor (\( \tau = RC \)). This is one time constant for these values.
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