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
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.