Question

If charge flown through a cross section of wire in one direction 0 to t is given by q=3 sin(3t) then Find out the area under i-t curve from t=(pi)/(9) to t=(pi)/(6) seconds :

Answer 1

Given:

The charge flown through a cross section of wire in one direction 0 to t is given by q=3 sin(3t).

To find:

The area under i-t curve from t=(pi)/(9) to t=(pi)/(6) seconds ?

Calculation:

We know that area under i-t graph gives us the net charge that flows through the conductor.

$\displaystyle \: area = \int_{ \frac{\pi}{9} }^{ \frac{\pi}{6} } \: i \: dt$

$\displaystyle \implies area = \bigg \{ q \bigg \}_{ \frac{\pi}{9} }^{ \frac{\pi}{6} }$

$\displaystyle \implies area = \bigg \{ 3 \sin(3t) \bigg \}_{ \frac{\pi}{9} }^{ \frac{\pi}{6} }$

$\displaystyle \implies area = 3 \{\sin(3 \times \frac{\pi}{6} ) - \sin(3 \times \frac{\pi}{9} ) \}$

$\displaystyle \implies area = 3 \{\sin( \frac{\pi}{2} ) - \sin(\frac{\pi}{3} ) \}$

$\displaystyle \implies area = 3 \{ \frac{1}{ \sqrt{2} } - \frac{1}{2} \}$

$\displaystyle \implies area = 3 \times 0.207$

$\displaystyle \implies area = 0.621$

So, area in that interval is 0.621.

Answer 2

Answer:

$Given:</p><p></p><p>The charge flown through a cross section of wire in one direction 0 to t is given by q=3 sin(3t).</p><p></p><p>To find:</p><p></p><p>The area under i-t curve from t=(pi)/(9) to t=(pi)/(6) seconds ?</p><p></p><p>Calculation:</p><p></p><p>We know that area under i-t graph gives us the net charge that flows through the conductor.</p><p></p><p>\displaystyle \: area = \int_{ \frac{\pi}{9} }^{ \frac{\pi}{6} } \: i \: dtarea=∫9π6πidt</p><p></p><p>\displaystyle \implies area = \bigg \{ q \bigg \}_{ \frac{\pi}{9} }^{ \frac{\pi}{6} }⟹area={q}9π6π</p><p></p><p>\displaystyle \implies area = \bigg \{ 3 \sin(3t) \bigg \}_{ \frac{\pi}{9} }^{ \frac{\pi}{6} }⟹area={3sin(3t)}9π6π</p><p></p><p>\displaystyle \implies area = 3 \{\sin(3 \times \frac{\pi}{6} ) - \sin(3 \times \frac{\pi}{9} ) \}⟹area=3{sin(3×6π)−sin(3×9π)}</p><p></p><p>\displaystyle \implies area = 3 \{\sin( \frac{\pi}{2} ) - \sin(\frac{\pi}{3} ) \}⟹area=3{sin(2π)−sin(3π)}</p><p></p><p>\displaystyle \implies area = 3 \{ \frac{1}{ \sqrt{2} } - \frac{1}{2} \}⟹area=3{21−21}</p><p></p><p>\displaystyle \implies area = 3 \times 0.207⟹area=3×0.207</p><p></p><p>\displaystyle \implies area = 0.621⟹area=0.621</p><p></p><p>So, area in that interval is 0.621.</p><p></p><p>$