Answer
(a) $R_2$
(b) $R_1$
Work Step by Step
(a) We know that when the two resistors are connected in series they will have the same current. Since $P=I^2R$, this shows that power dissipated is directly proportional to the resistance. Thus, resistor $R_2$ which has more resistance will dissipate more power.
(b) Since it is given that the resistors are connected in parallel, they will have the same voltage. For this combination of resistors $P=\frac{V^2}{R}$. This equation shows that power dissipated is inversely proportional to the resistance. Thus, resistor $R_1$ having less resistance will dissipate more power as compared to $R_2$.
