Solve RL Parallel Circuit: Find VL(0) Across 5ohm Resistor

[jendrix]

Homework Statement

Find the open circuit voltage VL(0) across the 5ohm resistor

Homework Equations

Z1*Z2/Z1+Z2

V(Z2*Z1+Z2)

The Attempt at a Solution

I thought for the open circuit voltage that due to infinite resistance of OC that the voltage across j35 inductor would be the equivalent of VL(0).

So i'd use the voltage divider equation across the 2 inductors in the left hand loop.

j35/(j35+j0.3)

Solving this using by multiplying the numerator and denominator by the conjugate of j35+j0.3

j35/(j35+j0.3)

Am I on the right track?

Thanks

 

[NascentOxygen]

If you have interpreted the question correctly, then you have started it the right way.

But I'm not convinced that the question is "Find the open circuit voltage VL(0) across the 5ohm resistor". Is that your wording?

Can you quote the instructions exactly?

 

[gneill]

<image snipped>

Homework Statement

Find the open circuit voltage VL(0) across the 5ohm resistor

Homework Equations

Z1*Z2/Z1+Z2

V(Z2*Z1+Z2) ------ Huh?

The Attempt at a Solution

I thought for the open circuit voltage that due to infinite resistance of OC that the voltage across j35 inductor would be the equivalent of VL(0).

So i'd use the voltage divider equation across the 2 inductors in the left hand loop.

j35/(j35+j0.3)

Solving this using by multiplying the numerator and denominator by the conjugate of j35+j0.3

j35/(j35+j0.3)

Am I on the right track?

Yup. Should be fine so long as the inductors are not coupled by mutual inductance.

 

[jendrix]

Sorry, it's two separate questions, first find the open circuit voltage VL(0) where there's an OC instead of the resistor.Then secondly find the voltage over the resistor.

Sorry, misquoted the potential divider equation, it should have been

V(Z2/Z1+Z2)

 

[jendrix]

Am I right in thinking that as it's just 2 inductors then the voltage across will be in phase with each other, hence

v*(j35/j35+j0.3) the j will cancel leaving 100*(35/35.3) =99.2v? For the OC voltage?

Thanks

 

[NascentOxygen]

Sorry, it's two separate questions, first find the open circuit voltage VL(0) where there's an OC instead of the resistor.Then secondly find the voltage over the resistor.

okay

Sorry, misquoted the potential divider equation, it should have been

V(Z2/Z1+Z2)

I'd not hesitate to mark that wrong.

 

[jendrix]

Sorry I'll try again

VL(0)=V*(Z2/(Z1+Z2))

=100*(j35/(j35+j0.3))

=100*(35/35.3)

VL(0)=99.15v

 

[NascentOxygen]

Sorry I'll try again

VL(0)=V*(Z2/(Z1+Z2))

=100*(j35/(j35+j0.3))

=100*(35/35.3)

VL(0)=99.15v

Looks okay. That's the easy part out of the way ...

 

[jendrix]

Thanks, so for VL would loop analysis work? Or should I be looking at using Z1*Z2/(Z1+Z2) to simplify the circuit?

Thanks

 

[NascentOxygen]

Thanks, so for VL would loop analysis work? Or should I be looking at using Z1*Z2/(Z1+Z2) to simplify the circuit?

Any valid method will work. Choose one, or be daring and try a few different ways!

 

[jendrix]

The aim was to get us practising with complex impedences.So my plan is to find overall impedence using

Z1*Z2/Z1+Z2

then add this to the 0.3 inductor closest to the voltage supply.This gave me 1.28j+4.82 which gives me the wrong VL compared to the answer provided (98.45v)

 

[NascentOxygen]

There seems to be a problem with the parentheses keys on your keyboard...?

You won't get the answer in one step. There are a couple of 'potential divider' steps to consider when determining VL.

If you don't provide your working, it is difficult to say whether you are going about it the right way or not.

 

[jendrix]

Sorry again :)

Thanks, I think I understand now, it's just the complex arithmitic I think I slipped up on.It'll be a struggle to type it all up on here, so I'll stick with it and look for a solution.

Thanks again for your assistance.

 

[NascentOxygen]

Sorry again :)

Thanks, I think I understand now, it's just the complex arithmitic I think I slipped up on.It'll be a struggle to type it all up on here, so I'll stick with it and look for a solution.

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