Solving Homework Eqns: R, C, P_ex, P_app

  • Thread starter Thread starter titanae
  • Start date Start date
  • Tags Tags
    Homework
Click For Summary
SUMMARY

The discussion focuses on solving two differential equations related to electrical circuits, specifically for variables R (resistance), C (capacitance), P_ex (external power), and P_app (applied power). The solutions for V_i(t) and V_e(t) are derived from the equations, with V_i(t) expressed as C(P_app - P_ex)(1 - e^(t/RC)) and V_e(t) as C(-P_ex)(e^(t/RC) - 1). The final equation relating P_ex to the other variables is established as P_ex = [((e^(t_i)/RC) - 1) * P_app] / ((e^(t_tot)/RC) - 1). The accuracy of the solutions A and B is questioned, particularly in the context of setting V_i equal to V_T for part C.

PREREQUISITES
  • Understanding of differential equations in electrical circuits
  • Familiarity with the concepts of resistance (R) and capacitance (C)
  • Knowledge of initial conditions in solving differential equations
  • Proficiency in exponential functions and their applications in circuit analysis
NEXT STEPS
  • Review the method of solving first-order linear differential equations
  • Study the application of Laplace transforms in circuit analysis
  • Explore the relationship between time constants and circuit behavior
  • Investigate the implications of initial conditions on circuit response
USEFUL FOR

Students and professionals in electrical engineering, particularly those focused on circuit analysis and differential equations in electrical systems.

titanae
Messages
2
Reaction score
0

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) ??
 
Physics news on Phys.org
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
 
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?
 

Similar threads

  • · Replies 37 ·
2
Replies
37
Views
8K
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
0
Views
1K
Replies
20
Views
2K
Replies
2
Views
2K