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Question 2 An electric circuit consists of a battery (voltage source) connected to a resistor and capacitor in series. The resulting ordinary differential equation for the voltage on the capacitor Vc(t) is as follows: We will use the values (b) Vin function dVc Vin dt RC = = R VR 1 RC satisfies the ODE and the initial condition. 1 1000 C= 0.001 Vo = 9 Here resistance R is in Ohms, time t is in seconds, capacitance C is in Farads, and voltage Vin is in Volts. Initially, the voltage on the capacitor is zero: Vc (0) = 0. (a) Assume the resistance is constant: R Vc(t). = Vc T O Vc(t)= 9(1e-10t) 100. Check by direct substitution that the Now assume that the resistor slowly degrades, so that its resistance increases with time: R = R(t) = 100 + 2t.

Question 2 An electric circuit consists of a battery (voltage source) connected to a resistor and capacitor in series. The resulting ordinary differential equation for the voltage on the capacitor Vc(t) is as follows: We will use the values (b) Vin function dVc Vin dt RC = = R VR 1 RC satisfies the ODE and the initial condition. 1 1000 C= 0.001 Vo = 9 Here resistance R is in Ohms, time t is in seconds, capacitance C is in Farads, and voltage Vin is in Volts. Initially, the voltage on the capacitor is zero: Vc (0) = 0. (a) Assume the resistance is constant: R Vc(t). = Vc T O Vc(t)= 9(1e-10t) 100. Check by direct substitution that the Now assume that the resistor slowly degrades, so that its resistance increases with time: R = R(t) = 100 + 2t.

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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Question 2 An electric circuit consists of a battery (voltage source) connected to a resistor and capacitor in series. The resulting ordinary differential equation for the voltage on the capacitor Vc(t) is as follows: We will use the values (b) Vin function dVc dt + - + C = 0.001 Vin 1 RC RC - R VR satisfies the ODE and the initial condition. 1 1000' Vo = 9 Here resistance R is in Ohms, time t is in seconds, capacitance C is in Farads, and voltage Vin is in Volts. Initially, the voltage on the capacitor is zero: Vc(0) = = 0. (a) Assume the resistance is constant: R Vc(t). = Vc Q Vc(t) = 9(1-e-10t) 100. Check by direct substitution that the Now assume that the resistor slowly degrades, so that its resistance increases with time: = R(t) = = 100 + 2t. R= Solve the ordinary differential equation for Vc(t) with this resistance, using either separation of variables, or the integrating factor method. Make sure to also include the initial condition.
Transcribed Image Text:Question 2 An electric circuit consists of a battery (voltage source) connected to a resistor and capacitor in series. The resulting ordinary differential equation for the voltage on the capacitor Vc(t) is as follows: We will use the values (b) Vin function dVc dt + - + C = 0.001 Vin 1 RC RC - R VR satisfies the ODE and the initial condition. 1 1000' Vo = 9 Here resistance R is in Ohms, time t is in seconds, capacitance C is in Farads, and voltage Vin is in Volts. Initially, the voltage on the capacitor is zero: Vc(0) = = 0. (a) Assume the resistance is constant: R Vc(t). = Vc Q Vc(t) = 9(1-e-10t) 100. Check by direct substitution that the Now assume that the resistor slowly degrades, so that its resistance increases with time: = R(t) = = 100 + 2t. R= Solve the ordinary differential equation for Vc(t) with this resistance, using either separation of variables, or the integrating factor method. Make sure to also include the initial condition.
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