Explanation Given information: The energy(in joules) expended by the bulb over a period of time T second is, U = ∫ 0 T P ( t ) d t , where P = V 2 R is power(J/s), R = 144 Ω and voltage varies as V = V 0 ( 2 π f t ) V 0 = 110 V , f = 60 H z Formula used: The integral formula ∫ cos x d x = sin x + C , where C is an arbitrary The trigonometric property cos 2 x = 1 − 2 sin 2 x sin 2 x = 1 − cos 2 x 2 Calculation: To compute U = ∫ 0 T P ( t ) d t , putting value of P(t) & then R U = ∫ 0 T P ( t ) d t = ∫ 0 T V 2 R d t = 1 144 ∫ 0 T V 2 d t , putting value of V and V 0 , thus U = 1 144 ∫ 0 T V 2 d t = 1 144 ∫ 0 T ( V 0 sin ( 2 π f t ) ) 2 d t = V 0 2 144 ∫ 0 T sin 2 ( 2 π f t ) d t = 110 2 144 ∫ 0 T sin 2 ( 2 π f t ) d t = 3025 36 ∫ 0 T sin 2 ( 2 π f t )

ACHIEVE FOR CALCULUS 1 TERM >CSI CUSTOM<

4th Edition

ISBN: 9781319411657

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 7.2, Problem 72E

To determine

To compute U if the bulb remains opens for 5 hours, where U is the energy (in joules) expended by the bulb over a period of time T second is,

U=∫0TP(t)dt, where P=V2R is power(J/s), R=144Ω and voltage varies as V=V0(2πft)

V0=110V,f=60Hz

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Chapter 7.2, Problem 71E

Chapter 7.2, Problem 73E

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Chapter 7 Solutions

ACHIEVE FOR CALCULUS 1 TERM >CSI CUSTOM<

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To compute U if the bulb remains opens for 5 hours, where U is the energy (in joules) expended by the bulb over a period of time T second is, U = ∫ 0 T P ( t ) d t , where P = V 2 R is power(J/s), R = 144 Ω and voltage varies as V = V 0 ( 2 π f t ) V 0 = 110 V , f = 60 H z

Solution Summary: The author calculates the energy (in joules) expended by the bulb over a period of time T second.

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To compute U if the bulb remains opens for 5 hours, where U is the energy (in joules) expended by the bulb over a period of time T second is,

U=∫0TP(t)dt, where P=V2R is power(J/s), R=144Ω and voltage varies as V=V0(2πft)

V0=110V,f=60Hz

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Chapter 7 Solutions

To compute U if the bulb remains opens for 5 hours, where U is the energy (in joules) expended by the bulb over a period of time T second is, U = ∫ 0 T P ( t ) d t , where P = V 2 R is power(J/s), R = 144 Ω and voltage varies as V = V 0 ( 2 π f t ) V 0 = 110 V , f = 60 H z

Solution Summary: The author calculates the energy (in joules) expended by the bulb over a period of time T second.

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To compute U if the bulb remains opens for 5 hours, where U is the energy (in joules) expended by the bulb over a period of time T second is, U=∫0TP(t)dt, where P=V2R is power(J/s), R=144Ω and voltage varies as V=V0(2πft) V0=110V,f=60Hz

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Chapter 7 Solutions

To compute U if the bulb remains opens for 5 hours, where U is the energy (in joules) expended by the bulb over a period of time T second is,

U=∫0TP(t)dt, where P=V2R is power(J/s), R=144Ω and voltage varies as V=V0(2πft)

V0=110V,f=60Hz