Question:
Two equal resistances when connected in series to a battery, consume electric power of $60 \mathrm{~W}$. If these resistance are now connected in parallel combination to the same battery, the electric power consumed will be :
Correct Option: , 2
Solution:
(2) When two resistances are connected in series,
$R_{e q}=2 R$
Power consumed, $P=\frac{\varepsilon^{2}}{R_{e q}}=\frac{\varepsilon^{2}}{2 R}$
In parallel condition, $R_{e q}=R / 2$.
New power, $\mathrm{P}^{\prime}=\frac{\varepsilon^{2}}{(\mathrm{R} / 2)}$
or $\mathrm{P}^{\prime}=4 \mathrm{P}=240 \mathrm{~W} \quad(\because \mathrm{P}=60 \mathrm{~W})$