Two identical resistors, each of resistance $2 \Omega$, are connected in turn : $(i)$ in series, and $(ii)$ in paralle to a battery of $12 ,V$. Calculate the ratio of power consumed in the two cases.
Two identical resistors, each of resistance $2 \Omega$, are connected in turn : $(i)$ in series, and $(ii)$ in paralle to a battery of $12 ,V$. Calculate the ratio of power consumed in the two cases.
Given $\quad R_{1}=2 \Omega, R_{2}=2 \Omega, V=12 V$
$(i)$ When resistors are connected in series, then
$R _{ S }=2+2=4 \Omega$
Therefore, power consumed
$P_{S}=\frac{V^{2}}{R_{S}}=\frac{(12)^{2}}{4}=36 W$
$(ii)$ When resistors are connected in parallel, then
$R_{P}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{2 \times 2}{2+2}=1 \Omega$
Therefore, power consumed
$P_{P}=\frac{V^{2}}{R_{P}}=\frac{(12)^{2}}{1}=144 W$
Hence, ratio $\frac{P_{P}}{P_{S}}=\frac{144}{36}=4$
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