College work ?

[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:

 

[R32]

Hey buddy, is that all that the question stated? No applied voltage or anything you've left out?

 

[mechelec]

are you sure the coil is shunted by a 0.005 ohm resistor

 

[kenny james]

try and find the answer yourself without asking!, you cannot take us in with you on your next exam, if you get it wrong no doubt the tutor will take you through it to explain the correct answer if you get it right a pat on your back.

 

[trev]

Post up what you think is the answer mate. I'd put money on it then being discussed in detail

 

[Bobster]

Re-type how the question was given to you word for word.

For a third year question I wouldn't see them giving you that question without an applied voltage.

 

[Knobhead]

49.9375A
Now work out how I did it.
It is "Homework"

 

[MarkieSparkie]

Current though 4 Ohm Coil = 62.4mA (rounded to 1 DP)

 

[Knobhead]

Ooopps as usual I misread the question! But Markie made me rethink, I was 0.00008A out in my original answer.
So the current for the 4 Ώ coil would be 62.421973mA.

So OK Markie and I agree on 62.42mA, but will we tell which way we did it? And I’ll be honest I used two methods because someone set seeds of doubt in my mind.

 

[Knobhead]

Eeerr Markie, we’re both wrong, decimal point in wrong place!

6.242mA!

 

[telectrix]

6.242mA here also

 

[telectrix]

6.242mA here also. a simple ratio calculation.

 

[Des 56]

The numbers are easy,its that damn decimal point that likes to take the P

 

[mechelec]

someone put me out of my misery, I get 62.422mA where am I going wrong?? (apart from sitting on here pondering problems that I dont need to ponder)

 

[Knobhead]

Move the decimal point and you've got the same answer. mili = 1/1000

I notice we’ve had no response from our OP

 

[MarkieSparkie]

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)

 

[Knobhead]

Current = 50A
Rc = 4 Ώ
Rs = 0.005 Ώ
Is = Shunt current
Rp = Parallel resistance

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

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!

 

[Knobhead]

Forgive me?

 

[mechelec]

Does that mean I was right for a change.

 

[Knobhead]

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.

 

[mechelec]

I suspect the OP has not come back because no one gave him an instant answer and told him how to do it although I could be wrong

 

[trev]

Mech you're such a cynic. The gadgie might be busy working the answer out

 

[Knobhead]

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.

 

[mechelec]

Mech you're such a cynic. The gadgie might be busy working the answer out

Age makes you cynical

 

[Lee Kerr]

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

 

[trev]

Age makes you cynical

Dead right mate, I used to be easy going