Question 22 of 190

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

A$3:2$B$3:1$C$3:4$D$4:3$

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Answer: (B) $3:1$

Heat produced in resistor when direct current flows is ${H}_{1}={i}^{2}{R}_{1}t$
${H}_{1}={4}^{2}\left(3\right)t=16\left(3\right)t$
Heat produced in resistor when alternating current flows is ${H}_{2}={i}_{rms}^{2}{R}_{2}t\left\{{i}_{rms}=\dfrac{{i}_{0}}{\sqrt{2}}\right\}$
${H}_{2}=\dfrac{{i}_{0}^{2}}{2}{R}_{2}t$
${H}_{2}=\dfrac{{4}^{2}}{2}\times \left(2\right)t=16t$
Thus, the ratio is ${H}_{1}:{H}_{2}=3:1$