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Power depends on voltage across the circuit and resistance of the circuit

\begin{equation} P = \frac{V^2}{R};\\ P_{series} = \frac{V^2}{(R_1+R_2)};\\ P_{parallel} = \frac{V^2}{(R_1^{-1}+R_2^{-1})}=\frac{V^2}{\frac{R_1+R_2}{R_1R_2}}=\frac{V^2R_1R_2}{R_1+R_2};\\ \frac{P_{series}}{P_{parallel}} = \frac{1}{R_1R_2} \end{equation}

i.e. the answer depends on the product of the two resistances.

i.e. if $R_1R_2 > 1$, $P_{series} < P_{parallel}$
if $R_1R_2 < 1$, $P_{series} > P_{parallel}$

Power depends on voltage across the circuit and resistance of the circuit

\begin{equation} P = \frac{V^2}{R};\\ P_{series} = \frac{V^2}{(R_1+R_2)};\\ P_{parallel} = \frac{V^2}{(R_1^{-1}+R_2^{-1})^{-1}}=\frac{V^2}{\frac{R_1R_2}{R_1+R_2}}=\frac{V^2(R_1+R_2)}{R_1R_2};\\ \frac{P_{series}}{P_{parallel}} = \frac{R_1R_2}{(R_1+R_2)^2} \end{equation}

Since $R_1$ and $R_2$ are always positive, $R_1R_2 < (R_1+R_2)^2$ i.e. $P_{series} < P_{parallel}$

Power depends on voltage across the circuit and resistance of the circuit

\begin{equation} P = \frac{V^2}{R};\\ P_{series} = \frac{V^2}{(R_1+R_2)};\\ P_{parallel} = \frac{V^2}{(R_1^{-1}+R_2^{-1})}=\frac{V^2}{\frac{R_1+R_2}{R_1R_2}}=\frac{V^2R_1R_2}{R_1+R_2};\\ \frac{P_{series}}{P_{parallel}} = \frac{1}{R_1R_2} \end{equation}

i.e. the answer depends on the product of the two resistances.

Power depends on voltage across the circuit and resistance of the circuit

\begin{equation} P = \frac{V^2}{R};\\ P_{series} = \frac{V^2}{(R_1+R_2)};\\ P_{parallel} = \frac{V^2}{(R_1^{-1}+R_2^{-1})}=\frac{V^2}{\frac{R_1+R_2}{R_1R_2}}=\frac{V^2R_1R_2}{R_1+R_2};\\ \frac{P_{series}}{P_{parallel}} = \frac{1}{R_1R_2} \end{equation}

i.e. the answer depends on the product of the two resistances.

i.e. if $R_1R_2 > 1$, $P_{series} < P_{parallel}$
if $R_1R_2 < 1$, $P_{series} > P_{parallel}$

Power depends on voltage across the circuit and resistance of the circuit

\begin{equation} P = VR;\\ P_{series} = V(R_1+R_2);\\ P_{parallel} = V(R_1^{-1}+R_2^{-1})=V\frac{R_1+R_2}{R_1R_2};\\ \frac{P_{series}}{P_{parallel}} = R_1R_2 \end{equation}

i.e. the answer depends on the product of the two resistances.

Power depends on voltage across the circuit and resistance of the circuit

\begin{equation} P = \frac{V^2}{R};\\ P_{series} = \frac{V^2}{(R_1+R_2)};\\ P_{parallel} = \frac{V^2}{(R_1^{-1}+R_2^{-1})}=\frac{V^2}{\frac{R_1+R_2}{R_1R_2}}=\frac{V^2R_1R_2}{R_1+R_2};\\ \frac{P_{series}}{P_{parallel}} = \frac{1}{R_1R_2} \end{equation}

i.e. the answer depends on the product of the two resistances.