Question

two resistors, r1 and r2, are connected in series to a 5.00 V battery. R1 is 12.0 Oh. a current of 0.167 A flows through the circuit. what is resistance of R2?
(hint: what is the total resistence in the circuit?)
(unit=Ohms)

Answer 1

The two formulas needed to answer this question is the ohm's law and resistance in a series:
\(V=IR\)
\(R_{eq}=\sum R_n=R_1+R_2+R_3...\)

You will need to work 'backward' to find \(R_2\)

You are given that current is 0.167 amp and it is a 5.00 V voltaic cell
\(5.00~V=(0.167~A)(R)\)
Using algebra....
\(R=29.94≈29.9~Ω~\text{(3 s.f.)}\)
Now this is \(R_{eq}\) and you must find \(R_2\)...
\(29.9=12.0~Ω+R_2\)
Now you can find \(R_2\)