The figure shows current varying with time. The magnitude of constant direct current for a particular time which would produce same heat in an identical resistance as produced by given time varying current in three seconds.

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$\frac{1}{3} A$ when passed for 15 seconds

1 A when passed for 15 seconds

$\frac{\sqrt{5}}{3} A$ when passed for 3 seconds

$\frac{\sqrt{5}}{9} A$ when passed for 3 seconds

answer is B, C.

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Detailed Solution

$\begin{array}{l} Heat = \int^{} i^{2} R d t \\ = \int_{0}^{1} {(t)}^{2} R d t + \int_{1}^{2} 1^{2} R d t + \int_{2}^{3} {(3 - t)}^{2} R d t = \frac{1}{3} R + R + \frac{R}{3} = \frac{5 R}{3} \\ (∵ e q u a t i o n o f s t r a i g h t f i r s t l i n e = y = m x + c \Rightarrow i = 1 (t) + 0 = t;) \\ e q u a t i o n o f s t r a i g h t s e c o n d l i n e = y = m x + c \Rightarrow i = 0 (t) + (- 1) = - 1; \\ e q u a t i o n o f s t r a i g h t t h i r d l i n e = y = m x + c \Rightarrow i = - 1 (t) + C; 0 = (- 1) 3 + C; C = 3 ∴ i = - 1 t + 3 \\ {Case 1 : I}_{1}^{2} R \times 3 = \frac{5 R}{3} \\ \Rightarrow {I_{1}}^{2} = \frac{5}{9} \\ \Rightarrow I_{1} = \frac{\sqrt{5}}{3} A \\ {Case 2 : I}_{2}^{2} R \times 15 = \frac{5 R}{3} \\ \Rightarrow I_{2} = \frac{1}{3} A \end{array}$