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Q.

A resistor of resistance 100 $Ω$ is connected to an AC source $ε = (12 V) \sin (250 {πs}^{- 1}) t$ . Find the energy dissipated as heat during t = 0 to t = 1.0ms.

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a

$2 .61 \times 10^{- 6} J$

b

$0 .61 \times 10^{4} J$

c

$0 .61 \times 10^{- 4} J$

d

$2 .61 \times 10^{- 4} J$

answer is C.

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Detailed Solution

$\begin{array}{l} H = \frac{V^{2}}{R} t, E_{0} = 12 V, ω = 250 π, R = 100 Ω \\ H = \int_{0}^{H} dH = \int_{0}^{10^{- 3}} \frac{E_{0}^{2} \sin^{2} ωt}{R} dt \\ = \frac{144}{100} \int_{0}^{10^{- 3}} \sin^{2} ωtdt = 1 .44 \int_{0}^{10^{- 3}} (\frac{1 - \cos 2 ωt}{2}) dt \\ = 2 .61 \times 10^{- 4} J \end{array}$

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