Two equal resistances when connected in series to a battery, consume electric power of $60 \mathrm{~W}$. If…
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 :
$60 \mathrm{~W}$
$240 \mathrm{~W}$
$120 \mathrm{~W}$
$30 \mathrm{~W}$
Solution
When two resistances are connected in series, $R_{e q}=2 R$
Power consumed, $\mathrm{P}=\frac{\varepsilon^{2}}{\mathrm{R}_{\mathrm{ea}}}=\frac{\varepsilon^{2}}{2 \mathrm{R}}$
In parallel condition, $\mathrm{R}_{\mathrm{eq}}=\mathrm{R} / 2$ New power, $\mathrm{P}^{\prime}=\frac{\varepsilon^{2}}{(\mathrm{R} / 2)}$
or $\mathrm{P}^{\prime}=4 \mathrm{P}=240 \mathrm{~W}(\because \mathrm{P}=60 \mathrm{~W})$