Calculate voltage drop across the inductor

[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!

 

[cnh1995]

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?

 

[james123]

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?

 

[cnh1995]

I believe it's 12volts?

It should be, but does your equation give 12V for t=0?

i=I(1-e^(-tR/L)

That is correct. So what is the equation for voltage across the inductor?

 

[james123]

i=V(1-e^(-tR/L) ??

 

[cnh1995]

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.

 

[james123]

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

 

[cnh1995]

Vr = IR^- tR / L ??

No.
Look up the v-i relationship for inductor.

 

[cnh1995]

V=Ve^-t*R/L ?

Yes, but you have used the same symbol V for two different voltages.

V=12e^(-2*330/0.2)

No. You need to put t=2*time constant in this equation.

 

[james123]

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?

 

[james123]

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?

 

[cnh1995]

V=12e^-(1.212*330/0.

2*0.606 milliseconds =

 

[cnh1995]

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.

 

[cnh1995]

V=1.9835 volts ??

Ok, I just did the calculations on my phone. I think you only calculated the red part.

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

 

[james123]

Just re-did them and get the same answer with the '12e' in front of the red part..

 

[cnh1995]

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.

 

[james123]

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 ?

 

[cnh1995]

V=1.6568 volts ?

Right.

 

[james123]

Brilliant! Thanks for your help and patience, it's much appreciated!