(a) For the RL circuit, the equation was: dI E – L dt IR=0 Show by direct substitution that I(t) = (1 – eî') is indeed a solution. (b) For the LC circuit, we found the equation to be: dQ 1 dt2 LC Show by direct substitution that Q(t) = A cos(wt + 4) is indeed a solution if w = 1/VLC. (c) For the LRC circuit, we found the equation to be: dQ RdQ 1 dt2 LCO=0 L dt Show by direct substitution that Q(t) = Ae-(R/2L)t cos(w't + ¢) is indeed a solution if 1 w = R2 %3D LC 4L2
(a) For the RL circuit, the equation was: dI E – L dt IR=0 Show by direct substitution that I(t) = (1 – eî') is indeed a solution. (b) For the LC circuit, we found the equation to be: dQ 1 dt2 LC Show by direct substitution that Q(t) = A cos(wt + 4) is indeed a solution if w = 1/VLC. (c) For the LRC circuit, we found the equation to be: dQ RdQ 1 dt2 LCO=0 L dt Show by direct substitution that Q(t) = Ae-(R/2L)t cos(w't + ¢) is indeed a solution if 1 w = R2 %3D LC 4L2
Related questions
Question
100%
Transcribed Image Text:Confirming our solutions. Let's confirm the solutions that we wrote down to the various differential
equations representing each of the different L circuits we studied.
(a) For the RL circuit, the equation was:
dI
E – L
dt
IR=0
Show by direct substitution that I(t) = % (1– e i') is indeed a solution.
(b) For the LC circuit, we found the equation to be:
dQ
LC
Show by direct substitution that Q(t) = A cos(wt + 4) is indeed a solution if w = 1/vLC.
1
= 0
dt2
(c) For the LRC circuit, we found the equation to be:
dQ
R dQ
L dt
Show by direct substitution that Q(t) = Ae-(R/2L)t cos(w't + ø) is indeed a solution if
1
= 0
dt2
1
R2
w'
|
LC
4L2
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps
