Solving a Series RLC Transient Analysis Problem

Junior
Messages
1
Reaction score
0
Let's say we have a free-source series RLC circuit. The differential equation that describes the behavior of the transient is given by [tex]Ri+L\frac{di}{dt}+V=0[/tex], where V is the potential difference across the capacitor. But I have trouble understanding the way this equation is developed. For example, suppose that we have the following situation:
circuit.png


But if I apply the loop rule, I get:

[tex]-V+Ri+L\frac{di}{dt}=0[/tex]

What is wrong here? In the book they apply the loop rule to the same circuit above, but with the polarity of the capacitor inverted. But doesn't the current leave the positive terminal of it?

Thanks in advance!
 
Engineering news on Phys.org
Junior said:
Let's say we have a free-source series RLC circuit.
I presume you mean a source free circuit.

Re the equation, they don't say in which direction they are measuring the PD across the capacitor. Perhaps they are measuring it as (potential of lower plate) minus (potential of upper plate), in which case the given equation would work.
 

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
44
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K