# Voltage Drops in Transient RL Circuits

1. May 3, 2013

### tranceical

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

2. May 3, 2013

### rude man

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.

3. May 4, 2013

### tranceical

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.

4. May 4, 2013

### Staff: Mentor

Hi tranceical! [Broken]

That's the voltage across the inductor.

The voltage across the resistor at that moment is 14-2.6v.

I follow you so far, but then you lose me here:
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: May 6, 2017
5. May 4, 2013

### tranceical

Hi Nascent,

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

6. May 4, 2013

### Staff: Mentor

So it appears ....

They mean the voltage across a resistor after 'x' amount of time. [Broken]

Last edited by a moderator: May 6, 2017
7. May 4, 2013

### Staff: Mentor

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. [Broken]

Last edited by a moderator: May 6, 2017
8. May 5, 2013

### tranceical

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!