Question

A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a

Answer 1

     Considering the unknown resistence as R and using the Ohm's First Law, we have:

$i= \frac{V}{R_{eq}} \\ 0.5= \frac{12}{R+10} \\ R+10=24 \\ R=14-Ohm$
 
     The equivalent resistence is given by the resistor series with the lamp resistence.

$R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}$

If you notice any mistake in my english, please let me know, because i am not native.