RL circuit time constant question

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The time constant for the given RL circuit with a 20V input, an 8.2kΩ resistor, and a 50mH inductor is calculated as T = L/R, resulting in 6.09 microseconds. To find the voltage across the inductor (VL) after 1.5 time constants, the correct formula is VL = E(1 - e^(-t/T)), where t/T is substituted with 1.5. The calculation yields VL = 4.46V, which is correct for when the circuit is switched off. However, the teacher's answer of 15.54V likely pertains to the scenario where the circuit is switched on at t=0.
supra_tt
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Hi,

Was hoping someone could help me.

I have the following RL circuit:

20V input
8.2kohm resistor and 50mH Inductor in series.


Determine time constant:

I answered with: T = L/R 50mH / 8.2kohm

= 6.09 uS

Is this correct?

The question I am stuck on is:

Determine the value of VL after 1.5 time constants?

I have the formula, VL = E ( e - t / T)

I can't work out what to put for the t? I tried putting in:

20 (e - 1.5 / (6.09uS) but can't be right?

If someone could help I would be much appreciated

Thanks!
 
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1.5 time constant means that that t/T=1.5. Substitute 1.5 for t/T into the formula of VL, but do not forget using parentheses.

ehild
 
Thanks for the quick response!

I am getting VL = 4.46V, I was wondering if you could please confirm this?

I am a little confused because the teacher has written the answer as VL = 15.54V

Does this mean I just need to take 4.46 from 20 to get the answer?

Thanks again
 
VL(t) is different if the source is switched on or off at t=0. You calculated VL after the battery is switched off. In case of switching on, VL=E(1-e-t/T).

ehild
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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