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Solve the Problem V(t)=160cos(120πt)\mathrm { V } ( \mathrm { t } ) = 160 \cos ( 120 \pi \mathrm { t } )

Question 183

Multiple Choice

Solve the problem.
-Suppose that a hair dryer operates on a voltage that can be represented by the relation V(t) =160cos(120πt) \mathrm { V } ( \mathrm { t } ) = 160 \cos ( 120 \pi \mathrm { t } ) and that it draws a current represented by the relation I(t) =8cos(120\mathrm { I } ( \mathrm { t } ) = 8 \cos ( 120 where tt is time measured in seconds. The power consumed by the appliance is P=VI\mathrm { P } = \mathrm { VI } . Graph the power in [0,0.04,0.01][ 0,0.04,0.01 ] by [200,2000,200][ - 200,2000,200 ] and use an identity to write the expression for the power in the form P=acos(kπt) +d\mathrm { P } = \mathrm { a } \cos ( \mathrm { k } \pi \mathrm { t } ) + \mathrm { d } , where a\mathrm { a } , k\mathrm { k } , and d\mathrm { d } are constants.


A) P=640cos(120πt) +640P = 640 \cos ( 120 \pi t ) + 640
[0,0.04,0.01][ 0,0.04,0.01 ] by [200,2000,200][ - 200,2000,200 ]
 Solve the problem. -Suppose that a hair dryer operates on a voltage that can be represented by the relation  \mathrm { V } ( \mathrm { t } )  = 160 \cos ( 120 \pi \mathrm { t } )   and that it draws a current represented by the relation  \mathrm { I } ( \mathrm { t } )  = 8 \cos ( 120  where  t  is time measured in seconds. The power consumed by the appliance is  \mathrm { P } = \mathrm { VI } . Graph the power in  [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]  and use an identity to write the expression for the power in the form  \mathrm { P } = \mathrm { a } \cos ( \mathrm { k } \pi \mathrm { t } )  + \mathrm { d } , where  \mathrm { a } ,  \mathrm { k } , and  \mathrm { d }  are constants. A)   P = 640 \cos ( 120 \pi t )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     B)   P = 640 \cos ( 240 \pi t )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     C)   \mathrm { P } = 1280 \cos ( 240 \pi \mathrm { t } )  + 1280   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     D)   \mathrm { P } = - 640 \cos ( 240 \pi \mathrm { t } )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]

B) P=640cos(240πt) +640P = 640 \cos ( 240 \pi t ) + 640
[0,0.04,0.01][ 0,0.04,0.01 ] by [200,2000,200][ - 200,2000,200 ]
 Solve the problem. -Suppose that a hair dryer operates on a voltage that can be represented by the relation  \mathrm { V } ( \mathrm { t } )  = 160 \cos ( 120 \pi \mathrm { t } )   and that it draws a current represented by the relation  \mathrm { I } ( \mathrm { t } )  = 8 \cos ( 120  where  t  is time measured in seconds. The power consumed by the appliance is  \mathrm { P } = \mathrm { VI } . Graph the power in  [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]  and use an identity to write the expression for the power in the form  \mathrm { P } = \mathrm { a } \cos ( \mathrm { k } \pi \mathrm { t } )  + \mathrm { d } , where  \mathrm { a } ,  \mathrm { k } , and  \mathrm { d }  are constants. A)   P = 640 \cos ( 120 \pi t )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     B)   P = 640 \cos ( 240 \pi t )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     C)   \mathrm { P } = 1280 \cos ( 240 \pi \mathrm { t } )  + 1280   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     D)   \mathrm { P } = - 640 \cos ( 240 \pi \mathrm { t } )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]

C) P=1280cos(240πt) +1280\mathrm { P } = 1280 \cos ( 240 \pi \mathrm { t } ) + 1280
[0,0.04,0.01][ 0,0.04,0.01 ] by [200,2000,200][ - 200,2000,200 ]
 Solve the problem. -Suppose that a hair dryer operates on a voltage that can be represented by the relation  \mathrm { V } ( \mathrm { t } )  = 160 \cos ( 120 \pi \mathrm { t } )   and that it draws a current represented by the relation  \mathrm { I } ( \mathrm { t } )  = 8 \cos ( 120  where  t  is time measured in seconds. The power consumed by the appliance is  \mathrm { P } = \mathrm { VI } . Graph the power in  [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]  and use an identity to write the expression for the power in the form  \mathrm { P } = \mathrm { a } \cos ( \mathrm { k } \pi \mathrm { t } )  + \mathrm { d } , where  \mathrm { a } ,  \mathrm { k } , and  \mathrm { d }  are constants. A)   P = 640 \cos ( 120 \pi t )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     B)   P = 640 \cos ( 240 \pi t )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     C)   \mathrm { P } = 1280 \cos ( 240 \pi \mathrm { t } )  + 1280   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     D)   \mathrm { P } = - 640 \cos ( 240 \pi \mathrm { t } )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]

D) P=640cos(240πt) +640\mathrm { P } = - 640 \cos ( 240 \pi \mathrm { t } ) + 640
[0,0.04,0.01][ 0,0.04,0.01 ] by [200,2000,200][ - 200,2000,200 ]
 Solve the problem. -Suppose that a hair dryer operates on a voltage that can be represented by the relation  \mathrm { V } ( \mathrm { t } )  = 160 \cos ( 120 \pi \mathrm { t } )   and that it draws a current represented by the relation  \mathrm { I } ( \mathrm { t } )  = 8 \cos ( 120  where  t  is time measured in seconds. The power consumed by the appliance is  \mathrm { P } = \mathrm { VI } . Graph the power in  [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]  and use an identity to write the expression for the power in the form  \mathrm { P } = \mathrm { a } \cos ( \mathrm { k } \pi \mathrm { t } )  + \mathrm { d } , where  \mathrm { a } ,  \mathrm { k } , and  \mathrm { d }  are constants. A)   P = 640 \cos ( 120 \pi t )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     B)   P = 640 \cos ( 240 \pi t )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     C)   \mathrm { P } = 1280 \cos ( 240 \pi \mathrm { t } )  + 1280   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]     D)   \mathrm { P } = - 640 \cos ( 240 \pi \mathrm { t } )  + 640   [ 0,0.04,0.01 ]  by  [ - 200,2000,200 ]

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