Solving Homework Eqns: R, C, P_ex, P_app

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In summary, the conversation discusses equations for voltage (V) in a circuit with a resistor (R), capacitor (C), and external power sources (P_ex and P_app). The first equation, EQ 1, is solved for V_i(t) with an initial condition of V_i(0) = 0, while the second equation, EQ 2, is solved for V_e(t) with an initial condition of V_e(t_i) = V_T. The solution for P_ex is shown to be dependent on the constants R, C, and the time values t_i and t_tot, as well as the external power source P_app and the initial voltage V_T.
  • #1
titanae
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Homework Statement


Eq 1: R(dV_i/dt) + (1/C)V_i + P_ex = P_app , 0 <= t <= t_i
Eq 2: R(dV_e/dt) + (1/C)V_e + P_ex = 0 , t_i <= t <= t_tot

A) Solve EQ 1 for V_i(t) with the initial condition V_i(0) = 0
B) Solve EQ 2 for V_e(t) with the initial condition V_e(t_i) = V_T
C) Using V_i(t_i) = V_T, show

P_ex = [((e^(t_i)/RC) - 1) * P_app] / ((e^(t_tot)/RC) - 1)

Homework Equations


V_i(0) = 0

V_e(t_i) = V_i(t_i) = V_T

V_e(tot) = 0

R, C, P_ex, P_app are constants


The Attempt at a Solution



A) V_i = C(P_app - P_ex)(1- (e^(t/RC)))

B) V_T = C(- P_ex)((e^(t/RC))-1)

C) ??
 
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  • #2
I haven't checked A and B, but don't you just set V_i equal to V_T for C? Since V_i = C(P_app - P_ex)(1- (e^(t/RC))) and V_T = C(- P_ex)((e^(t/RC))-1), C(P_app - P_ex)(1- (e^(t/RC)))=V_T = C(- P_ex)((e^(t/RC))-1). Then solve for P_ex
 
  • #3
Yes i tried that and this is what i got

(P_app - P_ex)(1 - e^(t_i/RC)) = (-P_ex)(e^(t_tot/RC) - 1)

which when expanded out cancels P_ex

maybe another part is wrong?
 

1. What are the basic principles of solving homework equations?

The basic principles of solving homework equations involve understanding the problem, identifying the relevant equations, and performing the necessary calculations to find the solution. It is important to carefully read and interpret the given information, use correct units, and check for errors in calculations.

2. How do I solve equations involving resistance (R)?

To solve equations involving resistance, you can use Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). So, if you know any two of these values, you can solve for the third one.

3. What is the role of capacitance (C) in solving equations?

Capacitance is a measure of a capacitor's ability to store charge. In solving equations involving capacitance, you need to use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. This formula allows you to calculate any of these three variables if the other two are known.

4. How does external power (Pex) affect equations?

External power, also known as power supplied to the system, can either increase or decrease the energy of a circuit. In solving equations, you need to account for both external power and power applied by the source (Papp). The net power of the system can be calculated by adding or subtracting these two values.

5. Can I use the same method to solve all equations involving R, C, Pex, and Papp?

While the basic principles remain the same, the method for solving equations may vary depending on the specific problem. Some equations may require rearranging variables, using logarithms, or applying other mathematical concepts. It is important to carefully analyze the given problem and choose the appropriate method for solving it.

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