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Step 1: Understand the Series Combination
When two equal resistances, each of resistance $R$, are connected in series, their equivalent resistance becomes $2R$. Let the battery have an EMF of $\varepsilon$. The power $P$ consumed in the series combination is given by
$P = \dfrac{\varepsilon^{2}}{2R}.$
It is given that $P = 60\,\text{W}$.
Step 2: Express the EMF in Terms of $R$
From $P = 60\,\text{W} = \dfrac{\varepsilon^{2}}{2R},$ we conclude:
\[
\varepsilon^{2} = 120R.
\]
Step 3: Understand the Parallel Combination
When the same two resistances are connected in parallel, the equivalent resistance $R_{\text{parallel}}$ becomes $\dfrac{R}{2}.$
Step 4: Calculate Power in the Parallel Combination
The power $P'$ in the parallel combination is
\[
P' = \dfrac{\varepsilon^{2}}{R/2}
= \dfrac{\varepsilon^{2}}{\tfrac{R}{2}}
= 2 \,\dfrac{\varepsilon^{2}}{R}.
\]
Using the fact that $\dfrac{\varepsilon^{2}}{2R} = 60\,\text{W},$ observe that
\[
\tfrac{\varepsilon^{2}}{R} = 2 \times \dfrac{\varepsilon^{2}}{2R} = 2 \times 60\,\text{W} = 120\,\text{W}.
\]
Hence,
\[
P' = 2 \times 120\,\text{W} = 240\,\text{W}.
\]
Step 5: Final Answer
Therefore, when the two resistances are connected in parallel to the same battery, the power consumed is $240\,\text{W}.$
