Voltage Drops in Transient RL Circuits

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Discussion Overview

The discussion revolves around understanding voltage drops in transient RL circuits, specifically focusing on the calculation of voltage across an inductor and the interpretation of the term "voltage drop." Participants explore the implications of time constants and the definitions of voltage in the context of circuit analysis.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents a circuit example with a 14V DC supply, a 200-ohm resistor, and a 0.6H inductor, seeking to calculate the voltage across the inductor after a specified time.
  • Another participant emphasizes the need for clarity on where the 'voltage drop' is defined and requests a circuit diagram to better understand the situation.
  • Initial conditions are discussed, noting that at t=0, the current and voltage across the resistor are zero, and the voltage across the inductor is equal to the supply voltage.
  • There is a discussion about the definition of voltage drop, with one participant suggesting that it implies the difference in voltage after a certain time, while others clarify that it refers to the voltage across a specific component at that time.
  • One participant expresses confusion regarding examples that do not perform a final calculation to determine the voltage drop, questioning the interpretation of voltage drop versus voltage at a given time.
  • A later reply discusses the terminology of voltage drop in relation to Kirchhoff's Voltage Law (KVL), suggesting that it serves as a descriptive tool in circuit analysis.

Areas of Agreement / Disagreement

Participants express differing interpretations of the term "voltage drop," leading to some confusion. There is no consensus on whether voltage drop should be calculated as a difference from the supply voltage or simply as the voltage across a component at a given time.

Contextual Notes

Participants note the importance of initial conditions and definitions in understanding voltage behavior in circuits. The discussion highlights the potential for varying interpretations of voltage drop in different contexts.

Who May Find This Useful

This discussion may be useful for students and practitioners in electrical engineering and physics who are grappling with concepts of voltage in transient circuits, particularly in RL and RC circuits.

tranceical
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Hi all,

I have a general query about voltage drops. Below is an example of the sort of question i am dealing with.

If we have:
a series circuit with a 14v dc supply, with a resistor of 200ohms and an inductor with inductance 0.6H

use the exponential equation to work out voltage across the inductor: Vl = (V)e-Rt/l

Q. after 2 time constants (0.005seconds) work out the voltage drop across the inductor?

then for this example i work out that the Voltage after 0.005seconds is 2.6v (1dp)

So surely it follows that 'voltage drop' is equal to 14-2.6v? In some transient examples I've seen they don't seem to do the final sum at the end, hence my confusion.

Thanks a lot for your time and help
 
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Depends on where the 'drop' is defined.

You need to show us your circuit including what happens to it at t = 0 plus initial conditions.
 
Initial conditions:

At t = 0 the initial value of current is zero and the initial voltage drop across Vr (resistor) is also zero, but will rise exponentially in time.

The voltage across the inductor at any instant is Vl = V - Vr, at t = 0 Vl = V and as time goes on Vl will fall exponentially to zero.


I have also been unsure with very similar 'voltage drops' on discharging capacitors through resistors.
 
tranceical said:
Hi all,

I have a general query about voltage drops. Below is an example of the sort of question i am dealing with.

If we have:
a series circuit with a 14v dc supply, with a resistor of 200ohms and an inductor with inductance 0.6H

use the exponential equation to work out voltage across the inductor: Vl = (V)e-Rt/l

Q. after 2 time constants (0.005seconds) work out the voltage drop across the inductor?

then for this example i work out that the Voltage after 0.005seconds is 2.6v (1dp)
Hi tranceical!

That's the voltage across the inductor.

So surely it follows that 'voltage drop' is equal to 14-2.6v?
The voltage across the resistor at that moment is 14-2.6v.

I follow you so far, but then you lose me here:
In some transient examples I've seen they don't seem to do the final sum at the end, hence my confusion.
I have no idea what you mean. "Voltage drop" doesn't imply any particular element, it's just another phrase for "potential difference" or voltage across some element.
 
Last edited by a moderator:
Hi Nascent,

Thanks for your reply and welcome :).
I think as you say my issue is with the definition of voltage drop.

If in a question i am asked for for the voltage drop across a resistor after 'x' amount of time, what do they mean?

the word drop implies to me that they want to know how much the voltage has dropped by after 'x'. In the example from my first post, after 0.005 seconds the voltage across the inductor is 2.6v...so in my mind the voltage drop would be 14-2.6v (as this is the amount it has 'dropped')In my first post about other examples not doing the final sum - what i mean is, in the example questions I've seen where they ask for voltage drop across a resistor/inductor etc, they stop after the exponential equation and call this the voltage drop.
In my mind they have calculated the 'voltage at x amount of time' not how much the voltage has dropped by.

I hope this makes sense, sorry for any confusion.

Thanks again
 
tranceical said:
I think as you say my issue is with the definition of voltage drop.
So it appears ...

If in a question i am asked for for the voltage drop across a resistor after 'x' amount of time, what do they mean?
They mean the voltage across a resistor after 'x' amount of time.
 
Last edited by a moderator:
Post-musing: Why speak of "voltage drop" when "voltage" alone will do? Well, you can think of KVL as "Around a loop, the sum of the voltage rises = the sum of the voltage drops." It's just more descriptive, is an aid to calculations, and that's what language is all about. Voltage rises are due to batteries, etc.

Oops, that happy reader icon is a deal larger than I was expecting.
 
Last edited by a moderator:
Thanks very much for your help, its much appreciated. It all makes sense now. You mentioned voltage rise due to a battery which is a great way of explaining voltage drop, I was clearly reading into it too much.

PS. I am definitely a happy reader now so the big icon is justified :)
Cheers!
 

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