In this circuit,when a certain current flows, | vedclass

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Question

(D) Let the total current flowing through the $2 \Omega$ resistor be $I$. This current splits into two parallel branches: one containing $(6+9) \Omega

Vedclass · Accepted Answer

$2.88$

Vedclass · Answer

(D) Let the total current flowing through the $2 \Omega$ resistor be $I$. This current splits into two parallel branches: one containing $(6+9) \Omega = 15 \Omega$ and the other containing $5 \Omega$.Using the current divider rule,the current $I_1$ through the $5 \Omega$ resistor is:$I_1 = I 	imes \frac{15}{15+5} = I 	imes \frac{15}{20} = \frac{3}{4} I$Heat produced in the $5 \Omega$ resistor is $H_1 = I_1^2 	imes 5 	imes t = (\frac{3}{4} I)^2 	imes 5 	imes t = \frac{9}{16} I^2 	imes 5 	imes t = \frac{45}{16} I^2 t = 4.05 \ J$.From this,$I^2 t = \frac{4.05 	imes 16}{45} = 0.09 	imes 16 = 1.44$.Heat produced in the $2 \Omega$ resistor is $H = I^2 	imes 2 	imes t = 2 	imes (I^2 t) = 2 	imes 1.44 = 2.88 \ J$.

In this circuit,when a certain current flows,the heat produced in the $5 \Omega$ resistor is $4.05 \ J$ in a time $t$. The heat produced in the $2 \Omega$ resistor in the same time interval is: (in $J$)

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