# 5 mA, 50 mV 10ohms, could be used to measure a current.

• Dec 11th 2006, 12:29 PM
turbod15b
5 mA, 50 mV 10ohms, could be used to measure a current.
how do i show the formulae used :confused:

can someone show me how to draw a diagram including componenet values plz

cheers.
• Dec 11th 2006, 01:01 PM
topsquark
Quote:

Originally Posted by turbod15b
how do i show the formulae used :confused:

can someone show me how to draw a diagram including componenet values plz

cheers.

It looks like you are trying to use Ohm's Law? I think you need to tell us what the actual question is so we can help you answer it. There's precious little information here!

-Dan
• Dec 11th 2006, 02:41 PM
turbod15b
show, with the aid of a simple hand drawn diagram including component values, how a meter movement labelled: 5mA, 50 mV, 10 ohms, could be used to measure a current of 1A in a circuit.

clearly show the formulae used, & working out required, to calculate the necessary component values

:confused:
• Dec 12th 2006, 03:05 AM
earboth
Quote:

Originally Posted by turbod15b
show, with the aid of a simple hand drawn diagram including component values, how a meter movement labelled: 5mA, 50 mV, 10 ohms, could be used to measure a current of 1A in a circuit.

clearly show the formulae used, & working out required, to calculate the necessary component values

:confused:

Hello,

to measure the current you have to use a shunt which divides the current into two branches: The smaller amount (here 5 mA) passes through the meter, the bigger amount (here: 995 mA) passes through the resistor. See attachment.

Formulae in a parallel circuit:
math]U_{total}=U_1=U_2=...[/tex]
math]I_{total}=I_1+I_2+...[/tex]
math]\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+.. .[/tex]

You can take the results for your problem from the sketch.

EB

PS:: I've made a mistake typing Latex code but I couldn't find it. So I send you this reply as plain text. Hope it is useful nevertheless.