In the following circuit,1 ,Omega resistor di | vedclass

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(A) In the given circuit,the $3 \,\Omega$ and $1 \,\Omega$ resistors are in series with a $10 \,V$ battery.The equivalent resistance of the circuit is

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$P$

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(A) In the given circuit,the $3 \,\Omega$ and $1 \,\Omega$ resistors are in series with a $10 \,V$ battery.The equivalent resistance of the circuit is $R_{eq} = 3 \,\Omega + 1 \,\Omega = 4 \,\Omega$.The current flowing through the circuit is $i = \frac{V}{R_{eq}} = \frac{10 \,V}{4 \,\Omega} = 2.5 \,A$.The power dissipated in the $1 \,\Omega$ resistor is $P = i^2 R = (2.5)^2 	imes 1 = 6.25 \,W$.When the $1 \,\Omega$ resistor is replaced by a $9 \,\Omega$ resistor,the new equivalent resistance becomes $R'_{eq} = 3 \,\Omega + 9 \,\Omega = 12 \,\Omega$.The new current in the circuit is $i' = \frac{10 \,V}{12 \,\Omega} = \frac{5}{6} \,A$.The power dissipated in the $9 \,\Omega$ resistor is $P' = (i')^2 	imes 9 = \left(\frac{5}{6}ight)^2 	imes 9 = \frac{25}{36} 	imes 9 = \frac{25}{4} \,W = 6.25 \,W$.Since $P = 6.25 \,W$ and $P' = 6.25 \,W$,we have $P' = P$.

In the following circuit,$1 \,\Omega$ resistor dissipates power $P$. If the resistor is replaced by $9 \,\Omega$,the power dissipated in it is

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