A direct current of 4,A and an alternating cu | vedclass

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(B) For direct current $(DC)$,the heat produced is given by $H_1 = I^2 R_1 t$.Given $I = 4\,A$ and $R_1 = 3\,\Omega$,we have:$H_1 = (4)^2 	imes 3 	i

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$3: 1$

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(B) For direct current $(DC)$,the heat produced is given by $H_1 = I^2 R_1 t$.Given $I = 4\,A$ and $R_1 = 3\,\Omega$,we have:$H_1 = (4)^2 	imes 3 	imes t = 16 	imes 3 	imes t = 48t$.For alternating current $(AC)$,the heat produced is given by $H_2 = I_{rms}^2 R_2 t$,where $I_{rms} = \frac{I_0}{\sqrt{2}}$.Given peak value $I_0 = 4\,A$ and $R_2 = 2\,\Omega$,we have:$I_{rms} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\,A$.$H_2 = (2\sqrt{2})^2 	imes 2 	imes t = 8 	imes 2 	imes t = 16t$.The ratio of heat produced is $\frac{H_1}{H_2} = \frac{48t}{16t} = \frac{3}{1}$.Thus,the ratio is $3: 1$.

List-$I$	List-$II$
$A$. $AC$ generator	$I$. Detects the presence of current in the circuit
$B$. Galvanometer	$II$. Converts mechanical energy into electrical energy
$C$. Transformer	$III$. Works on the principle of resonance in $AC$ circuit
$D$. Metal detector	$IV$. Changes an alternating voltage for smaller or greater value

$A$ direct current of $4\,A$ and an alternating current of peak value $4\,A$ flow through resistances of $3\,\Omega$ and $2\,\Omega$ respectively. The ratio of heat produced in the two resistances in the same interval of time will be.

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