Two identical resistors,each of resistance 2 | vedclass

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(B) Given: $R_{1} = 2 \Omega, R_{2} = 2 \Omega, V = 12 	ext{ V}$.$(i)$ When resistors are connected in series,the equivalent resistance is:$R_{S} = R

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(B) Given: $R_{1} = 2 \Omega, R_{2} = 2 \Omega, V = 12 	ext{ V}$.$(i)$ When resistors are connected in series,the equivalent resistance is:$R_{S} = R_{1} + R_{2} = 2 + 2 = 4 \Omega$.Power consumed in series is:$P_{S} = \frac{V^{2}}{R_{S}} = \frac{12^{2}}{4} = \frac{144}{4} = 36 	ext{ W}$.$(ii)$ When resistors are connected in parallel,the equivalent resistance is:$R_{P} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{2 	imes 2}{2 + 2} = \frac{4}{4} = 1 \Omega$.Power consumed in parallel is:$P_{P} = \frac{V^{2}}{R_{P}} = \frac{12^{2}}{1} = 144 	ext{ W}$.The ratio of power consumed in the two cases (parallel to series) is:$\frac{P_{P}}{P_{S}} = \frac{144}{36} = 4:1$.

Sample	Resistivity $(\Omega \cdot m)$
$A$	$1.6 \times 10^{-8}$
$B$	$7.5 \times 10^{17}$
$C$	$44 \times 10^{-6}$

Two identical resistors,each of resistance $2 \Omega$,are connected in turn: $(i)$ in series,and $(ii)$ in parallel to a battery of $12 \text{ V}$. Calculate the ratio of power consumed in the two cases.

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