Jim Ortiz

[Jim Ortiz]

thank you, then I misunderstood the question now it makes sense.Sometimes it happens thats why we are not perfect…LOL

[Jim Ortiz]

yes thank you

[Jim Ortiz]

thank you

[Jim Ortiz]

that it is open circuit

[Jim Ortiz]

no voltage drop ?

[Jim Ortiz]

Susan
do I need to retake the test ?

[Jim Ortiz]

definitely I will, thank you again

[Jim Ortiz]

I understand it now, your right I didnt have to do the calculation ,as I WENT THROUGH my notes I seen where I made my mistakes I=E/R for current not E=I*R
The calculation comes out much different

[Jim Ortiz]

Thats okay I just wanted to be cure and yes your right I needed to ues the (I) for current,the more I keep going over it the more I understand what you are telling me .thank you

[Jim Ortiz]

ok so to do voltage drop the formula is I=E/R NOT E=I*R Is that right

[Jim Ortiz]

I have to leave I will respond when i get back thanks Im not ignoring you

[Jim Ortiz]

you add R1+R2=5005
you % 5005%120= 42
R1 you* 42*5=210
R2 you* 42*5000=210,000
is this right?

[Jim Ortiz]

I though it meant thousand?

[Jim Ortiz]

Fist of all sorry I did not respond back. The detector is close it is shunting the current around the main coil. I remember in a unit that I read current will take the path with no resistance it will go to neutral …

[Jim Ortiz]

module 6 unit 3

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