Question

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

• A. $60W$
• B. $240W$
• C. $30W$
• D. $120W$

Answer 1

The correct option is B

$240W$

Step 1: Given;

Power, $P_0=60W$

Let $P_1,P_2$ be the individual electrical powers of the two resistance respectively.

Also, $P_1=P_2$ since $P=I^{2}R$ and the resistances are equal and the current is also the same through both the resistors irrespective of the combination in which they are connected.

Step 2: Formula used:

In series combination, power is $P_0=\frac{P_1{P}_2}{P_1+P_2}$

Step 3: Find power when connected in parallel combination;

$60=\frac{P_1{P}_2}{P_1+P_2}$

$\Rightarrow 60=\frac{{P_1}^2}{2P_1}$ ($P_1=P_2$)

$$\begin{gathered} \Rightarrow \frac{P_1}{2}=60 \\ \Rightarrow P_1=120 \end{gathered}$$

Thus, $P_1=P_2=120W$

In parallel combination power is $P=P_1+P_2=120+120=240W$.

Hence, if the resistance is connected in parallel combination to the battery, then the electric power consumed will be $240W$.

Thus, option B is correct.