Solving for Vr and VL in an RLC Circuit

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SUMMARY

The discussion focuses on solving for the voltages across a resistor (Vr) and an inductor (VL) in an RLC circuit. The correct formulas are established: Vr = IR and VL = L(dI/dt), where L is the inductance and dI/dt is the rate of change of current. Participants clarify the importance of time dependence in voltage calculations and emphasize the need for accurate graphical representations of voltage over time. The final conclusion is that the voltage across the inductor is not simply a numerical value but is dependent on the current's time derivative.

PREREQUISITES
  • Understanding of Ohm's Law (U = IR)
  • Familiarity with inductance and Faraday's law
  • Knowledge of voltage-current relationships in RLC circuits
  • Ability to interpret and plot graphs of current and voltage over time
NEXT STEPS
  • Study the relationship between current and voltage in RLC circuits using simulation tools like LTspice.
  • Learn about the time-domain analysis of RLC circuits, focusing on transient responses.
  • Explore the concept of impedance in AC circuits and its applications.
  • Review the mathematical derivation of voltage across capacitors and inductors in dynamic conditions.
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing RLC circuits and their behavior in AC systems.

k31453
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Homework Statement



Hi i got this question ?



have to find Vr and VL

so can i use this formula :

VR = IR
VL = I * XL?


confuse ?? need help ??

am i on right track?
 
Last edited by a moderator:
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Impedance is used for AC circuits, when the current and voltage have sinusoidal time dependence. Here the time dependence of the current is different from a sine function. Go back to the definition of the voltage across a capacitor (in terms of charge) and inductor (in terms of derivative of current).

ehild
 
ehild said:
Impedance is used for AC circuits, when the current and voltage have sinusoidal time dependence. Here the time dependence of the current is different from a sine function. Go back to the definition of the voltage across a capacitor (in terms of charge) and inductor (in terms of derivative of current).

ehild


so answer is 0.002 * (20/1) = 0.04 for Vl

right?
 
k31453 said:
so answer is 0.002 * (20/1) = 0.04 for Vl

right?

? You have to show the time dependence of voltages. The answer is not a number.


ehild
 
ehild said:
? You have to show the time dependence of voltages. The answer is not a number.


ehild


whaaattt?

are you kidding mee !

need help please !
 
I try to help and I am not kidding. Read the problem text, please. ehild
 
ehild said:
I try to help and I am not kidding. Read the problem text, please.


ehild

yeah i know its time dependency .. this is the graph for voltage in inductor vs time in ms

 
Last edited by a moderator:
You sent the plot of voltage across the resistor. The shape is correct, but watch out the magnitude. The current is given in A (ampers) The resistance is 2Ω. What is the maximum voltage across the resistor?
What about the inductor?

ehild
 
ehild said:
You sent the plot of voltage across the resistor. The shape is correct, but watch out the magnitude. The current is given in A (ampers) The resistance is 2Ω. What is the maximum voltage across the resistor?
What about the inductor?

ehild



Got it right?
 
Last edited by a moderator:
  • #10
No, your diagrams are not correct.

Can you write the general expressions (definitions) relating voltage and current for a resistor and an inductor?
 
  • #11
k31453 said:
Got it right?

No. The maximum voltage for the resistor is about right now, but the shape is not. You remember Ohm's Law: The voltage across the resistor is U=IR, proportional to the current. The shape of the U(t) function follows the shape of I(t).

As for the inductor, remember Faraday's law about induced emf in a coil and inductance. How was it defined?

ehild
 
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  • #12
I know deal with resistor but i have no clue about inductor
 
  • #13
The voltage across the inductor is U=L(dI/dt) (it is proportional to the time derivative of the current). Determine the derivatives at the different sections and multiply by L.

ehild
 
  • #14
Along with U=L(dI/dt) , We should point out that the voltage across the capacitor v = (1/C) integral (i(t)) and of course voltage across the resistor is v = IR. The plot of current is simple enough that taking the integral or the derivative is simple.
 
  • #15
k31453 said:
so answer is 0.002 * (20/1) = 0.04 for Vl

right?

I agree with you. If the graph is current against time then this is the voltage across the inductor for the first part of the graph AND the last part of the graph. This voltage will be constant over these time intervals.
For the middle part it will be -0.04V.
For the flat bits of the graph there is no change of current with time...what will the voltage across the inductor be ??
 
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  • #16
technician said:
I agree with you. If the graph is current against time then this is the voltage across the inductor for the first part of the graph AND the last part of the graph. This voltage will be constant over these time intervals.
For the middle part it will be -0.04V.
For the flat bits of the graph there is no change of current with time...what will the voltage across the inductor be ??

The current changes 2 A in 1 ms,( Edit:20 A in 1 ms) so the voltage on the inductor is not 0.04 V.

ehild
 
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  • #17
It looks to me like 20A in 1ms ...
 
  • #18
technician said:
It looks to me like 20A in 1ms ...

I lost a zero, thanks. I edit the previous post. But still, UL is not right.

ehild
 
  • #19
ehild said:
The current changes 2 A in 1 ms,( Edit:20 A in 1 ms) so the voltage on the inductor is not 0.04 V.

ehild

so it will be this graph right because di/dt is derivitve !

 
Last edited by a moderator:
  • #20
k31453 said:
so it will be this graph right because di/dt is derivitve !

Excellent! Good solution, nice picture. (Only the unit V is missing from the vertical axis.)

ehild
 
Last edited by a moderator:
  • #21
ehild said:
Excellent! Good solution, nice picture. (Only the unit V is missing from the vertical axis.)

ehild

Yeah i solve the question but i forget how i did it and why is like that !?

:(:cry:
 
  • #22
Go to post #13:smile:

ehild
 
  • #23
ehild said:
Go to post #13:smile:

ehild

gotchya for 1 ms liner becomes flat and than flat becomes no gradient follow on right?
 
  • #24
The flat part has zero gradient so zero induced voltages.

ehild
 
  • #25
ehild said:
The flat part has zero gradient so zero induced voltages.

ehild

Nice ! thanks for all the help !
 
  • #26
You are welcome.:smile:

ehild
 

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