Calculate voltage drop across the inductor

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Discussion Overview

The discussion revolves around calculating the voltage drop across an inductor in a circuit consisting of a resistor and a DC supply. Participants explore the application of relevant equations and the implications of time constants in the context of an inductor's behavior in a series circuit.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant attempts to calculate the voltage drop across the inductor after two time constants using the equation V=V*(1-e^(-tR/L)), but expresses uncertainty about the application of the equation.
  • Another participant questions the initial conditions of the equation and whether the correct equation has been used for the inductor's voltage at t=0.
  • There is a suggestion that the current equation i=I(1-e^(-tR/L)) is also relevant, but confusion arises as it seems to yield the same current value repeatedly.
  • Participants discuss the need for the correct voltage equation for the inductor and the v-i relationship, indicating a lack of clarity on how to derive the voltage from the current equation.
  • One participant calculates the time constant and attempts to apply it to the voltage equation, leading to further confusion about the correct values to use.
  • There are multiple attempts to calculate the voltage drop, with varying results, and participants express frustration over repeated calculations yielding different outcomes.
  • Eventually, a participant arrives at a voltage of approximately 1.6568 volts after recalculating, but there is still uncertainty about the correctness of this value.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct application of the equations or the final voltage drop value, with ongoing confusion and multiple competing calculations presented throughout the discussion.

Contextual Notes

There are limitations in the clarity of the equations used, as well as potential misunderstandings regarding the definitions of voltage and current in relation to the inductor. Participants express uncertainty about the correct application of time constants and the implications for their calculations.

Who May Find This Useful

This discussion may be useful for students studying circuit theory, particularly those interested in the behavior of inductors in DC circuits and the application of time constants in electrical engineering problems.

james123

Homework Statement


An inductor of negligible resistance and an inductance of 0.2 H is connected in series with a 330 Ω resistor to a 12V d.c. supply. Determine:

  1. (b) the voltage drop across the inductor after two time constants
  2. (c) the voltage drop across the resistor after three time constants

Homework Equations



V=V*(1-e^(-tR/L))

The Attempt at a Solution


[/B]
I=V/R
I=12/330
I=0.036A

1. (b)
V=12*(1-e^(-2*330/0.2))
V=12V (obviously wrong)

Hi there, I think I might be a bit off with utilising this equation, I feel like I'm not plugging the values correctly/not using the correct values!

Any direction as to where I'm going wrong would be much appreciated!

Many thanks!
 
Last edited by a moderator:
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james123 said:
V=V*(1-e^(-tR/L))
According to this equation, what is the inductor voltage at t=0?
Are you sure you have used the correct equation?
 
Hi, thanks for replying.

I believe it's 12volts?

Have also seen this equation:
i=I(1-e^(-tR/L)

which would give me:

I=0.036(1-e^(-2*330/0.2))

But this just gives me I=0.036 again?
 
james123 said:
I believe it's 12volts?
It should be, but does your equation give 12V for t=0?
james123 said:
i=I(1-e^(-tR/L)
That is correct. So what is the equation for voltage across the inductor?
 
i=V(1-e^(-tR/L) ??
 
james123 said:
i=V(1-e^(-tR/L) ??
No, you are equating current with voltage. You have the correct equation for current. You need the equation for voltage across the inductor.
What is the general formula for voltage across an inductor? Look up v-i relation for inductor.
 
Okay, I've already calculated the time constant with

T=L/R = 0.2/330 = 0.0006 S
= 0.606 milliseconds

Just tried it this way too and no luck. Am along the right lines?

V=Ve^-t*R/L ?
V=12e^(-2*330/0.2)
V=12e^(-2*1650)
V=0v
 
Last edited by a moderator:
james123 said:
Vr = IR^- tR / L ??
No.
Look up the v-i relationship for inductor.
 
james123 said:
V=Ve^-t*R/L ?
Yes, but you have used the same symbol V for two different voltages.

james123 said:
V=12e^(-2*330/0.2)
No. You need to put t=2*time constant in this equation.
 
  • #10
Okay,

t=2:
2*0.606 milliseconds = 1.212

So,

V=12e^-(1.212*330/0.2)
V=0volts ??

Seem to keep going around in circles with this.

What am I plugging incorrectly?
 
  • #11
If I use the same equation but I use the time constant in seconds instead of milliseconds I get:

V=12e^-(0.0012*330/0.2)
V=1.9835 volts ??

Does this look right?
 
  • #12
james123 said:
V=12e^-(1.212*330/0.

james123 said:
2*0.606 milliseconds =
 
  • #13
james123 said:
If I use the same equation but I use the time constant in seconds instead of milliseconds I get:

V=12e^-(0.0012*330/0.2)
V=1.9835 volts ??

Does this look right?
I can't check the final answer right now, but you have plugged in the correct numbers.
 
  • #14
james123 said:
V=1.9835 volts ??
Ok, I just did the calculations on my phone. I think you only calculated the red part.
james123 said:
V=12e^-(0.0012*330/0.2)
 
  • #15
Just re-did them and get the same answer with the '12e' in front of the red part..
 
  • #16
james123 said:
Just re-did them and get the same answer with the '12e' in front of the red part..
That's not correct. Check your calculations.
 
  • #17
Okay I think I have it.

I separately calculated 0.0012*330/0.2=1.98

So,

V=12e^-1.98
V=1.6568 volts ?
 
  • #18
james123 said:
V=1.6568 volts ?
Right.
 
  • Like
Likes   Reactions: james123
  • #19
Brilliant! Thanks for your help and patience, it's much appreciated!
 

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