RLC Circuit

  • Thread starter freshbox
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  • #1
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Reference to the below 1st picture, I found that I = 0.916A.
I am trying to find I1/I2 and I tried to used the Current Divider Rule (2nd picture) But I am unable to solve it. May I know why?

Thanks.
 

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  • #2
gneill
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Reference to the below 1st picture, I found that I = 0.916A.
I am trying to find I1/I2 and I tried to used the Current Divider Rule (2nd picture) But I am unable to solve it. May I know why?

Thanks.
You'll have to show your attempt so that we can see how to help.

The first image is very dark; What's the value of the battery voltage? Is it 11 V?
 
  • #3
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I'm sorry. Yes the battery voltage is 11v.

My working:
Since I=0.916A

Using current divider rule to find i1:
I1=I x R3/R2+R3
I1=0.916 x 2
=0.916 x 2
=1.832A (Wrong Ans)

However the actual answer is 0.46A

So I'm wondering how come I cannot use CDR to solve for i1 or i2.

Please advise, thanks.
 
  • #4
gneill
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I'm sorry. Yes the battery voltage is 11v.

My working:
Since I=0.916A

Using current divider rule to find i1:
I1=I x R3/(R2+R3) <--- use parentheses to make operations clear!
I1=0.916 x 2 <---- How did you arrive at 2? What are R2 and R3?
Your approach is okay, but I think you've slipped up on plugging the correct values into the current divider equation.
 
  • #5
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I suspect that there is an inductor (L2) below resistor (R2) that's why I cannot use CDR.
 
  • #6
gneill
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I suspect that there is an inductor (L2) below resistor (R2) that's why I cannot use CDR.
Nope. No problem there; the inductor "disappears" at steady state. The problem seem to be what you're plugging in for R2 and R3, or if not that, what value you arrive at when you do the math.
 
  • #7
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Yea sorry I added my Resistor incorrectly, thanks for pointing out, I got the answer already.

I have another question to ask you: Is the voltage for the Capacitor is 10v or some of the voltage are absorb by by the Resistor?

To my understanding, at 5T (Steady State), Capacitor acts like an open circuit hence there is no current but it's voltage is at max value.

But I'm not sure whether the Capacitor is having full voltage or some of the voltage is absorb by the R (2k ohm)

Thank you.
 

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  • #8
gneill
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Yea sorry I added my Resistor incorrectly, thanks for pointing out, I got the answer already.

I have another question to ask you: Is the voltage for the Capacitor is 10v or some of the voltage are absorb by by the Resistor?

To my understanding, at 5T (Steady State), Capacitor acts like an open circuit hence there is no current but it's voltage is at max value.

But I'm not sure whether the Capacitor is having full voltage or some of the voltage is absorb by the R (2k ohm)

Thank you.
After a long time (5T and more), the current will go to zero. What's the voltage drop across the resistor if the current is zero?
 
  • #9
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0v. Capacitor = 10v
 

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