- #1

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## Homework Statement

Hi i got this question ?

[Broken]

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?

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- Thread starter k31453
- Start date

- #1

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Hi i got this question ?

[Broken]

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:

- #2

ehild

Homework Helper

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ehild

- #3

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ehild

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

right?

- #4

ehild

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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

- #5

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

ehild

whaaattt???

are you kidding mee !!

need help please !!!

- #6

ehild

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I try to help and I am not kidding. Read the problem text, please.

ehild

ehild

- #7

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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

[Broken]

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- #8

ehild

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What about the inductor?

ehild

- #9

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What about the inductor?

ehild

[Broken]

Got it right?

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- #10

gneill

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Can you write the general expressions (definitions) relating voltage and current for a resistor and an inductor?

- #11

ehild

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[Broken]

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

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I know deal with resistor but i have no clue about inductor

- #13

ehild

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ehild

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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

ehild

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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

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It looks to me like 20A in 1ms .....

- #18

ehild

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It looks to me like 20A in 1ms .....

I lost a zero, thanks. I edit the previous post. But still, U

ehild

- #19

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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 !!!

[Broken]

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- #20

ehild

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

[Broken]

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

ehild

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- #21

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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 !!???

:(

- #22

ehild

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Go to post #13

ehild

ehild

- #23

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Go to post #13

ehild

gotchya for 1 ms liner becomes flat and than flat becomes no gradient follow on right?

- #24

ehild

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The flat part has zero gradient so zero induced voltages.

ehild

ehild

- #25

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The flat part has zero gradient so zero induced voltages.

ehild

Nice !!! thanks for all the help !!

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