College work ? | Page 2 | on ElectriciansForums

Discuss College work ? in the Australia area at ElectriciansForums.net

L

Lee Kerr

Hey Guys
im a 3rd year apprentice struggling with my college work anyone got a clue in this question
HELP ME! please

A coil of resistance 4ohms is connected across a shunt of resistance 0.005ohms if the current in the main circuit is 50a what is the current of the coil ? :sick:
 
I don't think so! I'll stick with my previous answer.
Current Division Ratio
n = (R[SUB]shunt [/SUB]+ R[SUB]coil[/SUB]) / R[SUB]shunt[/SUB]
= (4 + 0.005) Ω / 0.005 Ω = 801
Coil Current
I[SUB]coil [/SUB]= I[SUB]supply[/SUB] / n
= 50A / 801
= 6.242197253 x 10[SUP]-2[/SUP] A
= 62.42 mA (62.4mA to 1 DP my previous answer QED)
 
Last edited by a moderator:
Current = 50A
Rc = 4 ÎŹ
Rs = 0.005 ÎŹ
Is = Shunt current
Rp = Parallel resistance

1/(Rc+Rs) = Rp
1/(4+0.005) = 0.004994 ÎŹ

Rp*I = V
0.004994 *50 = 0.249688V

V/Rs = Is
0.249688/4 = 0.062422A

For mA
0.062422*1000 = 62.422mA

Is = 62.422mA

I hold my hand up Markie, I was wrong!

:oops:
 
Last edited by a moderator:
Looks like another slice of umble pie for me.
Yes your right, I'm wrong. Bloody decimal points!
Can I have chips and gravy with the pie.

One thing it does show is there’s more that one way with questions to arrive at the same answer.

I notice our OP hasn’t come back.
 
Last edited by a moderator:
Can’t think why, he’s got both Markies and my answers.
You can’t get better proof than two different calculations with the same answer.
But he really should do his own homework.
 
Thanks for all the help guys this question has blown my brain to bits hahaha i get all of the working out but how did u know how to do it

(Rc+Rs)/Rs=N
(4+0.005)/0.005=801
Ic = Is/N
50/801 = 0.0624
0.0624 x 1000 = 62.4mA
 

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