Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistance are now connected in parallel combination to the same battery, the electric power consumed will be:

Let the resistance of each resistor be $$R$$ ohms. The battery maintains a constant potential difference, say $$V$$ volts.

First, we use the relation between power, voltage and resistance for a resistor:

$$P \;=\;\dfrac{V^2}{R_{\text{eq}}}$$

where $$P$$ is the electric power and $$R_{\text{eq}}$$ is the equivalent resistance connected across the battery.

We have the two resistances in series initially, so their equivalent resistance is

$$R_{\text{series}} \;=\; R + R \;=\; 2R.$$

The power consumed in this series arrangement is given to be 60 W. Hence,

$$P_{\text{series}} \;=\; 60 \;=\;\dfrac{V^2}{R_{\text{series}}} \;=\;\dfrac{V^2}{2R}.$$

Multiplying both sides by $$2R$$, we get

$$V^2 \;=\; 60 \times 2R \;=\; 120R.$$

Now the same two resistances are connected in parallel. The equivalent resistance of two equal resistors in parallel is

$$R_{\text{parallel}} \;=\;\dfrac{R}{2}.$$

The power consumed in the parallel arrangement is therefore

$$P_{\text{parallel}} \;=\;\dfrac{V^2}{R_{\text{parallel}}} \;=\;\dfrac{V^2}{\dfrac{R}{2}} \;=\; V^2 \times \dfrac{2}{R} \;=\; \dfrac{2V^2}{R}.$$

We substitute the value of $$V^2$$ obtained earlier, $$V^2 = 120R$$:

$$P_{\text{parallel}} \;=\;\dfrac{2 \times 120R}{R} \;=\; 240 \text{ W}.$$

Hence, the correct answer is Option B.