Solve the problem.
-Suppose that for an electrical appliance, voltage is given by $\mathrm { V } = 189 \sin 120 \pi \mathrm { t }$ and amperage by $I = 1.07 \sin 120 \pi \mathrm { t }$, where $t$ is time in seconds. Use identities to write the wattage $W = V I$ in the form $\mathrm { W } = \mathrm { a } \cos \omega \mathrm { t } + \mathrm { c }$.
A) $W = - 101.115 \cos 240 \pi t + 101.115$
B) $\mathrm { W } = - 101.115 \cos 120 \pi t + 101.115$
C) $\mathrm { W } = - 202.23 \cos 240 \pi \mathrm { t } + 202.23$
D) $W = 101.115 \cos 240 \pi t - 101.115$

Question

Quizplus · Accepted Answer

To find the wattage $W = V \times I$, we multiply the given expressions for voltage $V = 189 \sin 120 \pi t$ and amperage $I = 1.07 \sin 120 \pi t$. Using the trigonometric identity $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$, we get:$$W = 189 \times 1.07 \times \sin(120 \pi t) \sin(120 \pi t) = 202.23 \times \frac{1}{2}[\cos(0) - \cos(240 \pi t)]$$$$W = 101.115[1 - \cos(240 \pi t)] = 101.115 - 101.115\cos(240 \pi t)$$Rearranging to match the given form $W = a \cos \omega t + c$, we get:$$W = - 101.115 \cos(240 \pi t) + 101.115$$Therefore, the correct answer is A) $W = - 101.115 \cos 240 \pi t + 101.115$.

Solve the Problem V=189sin⁡120πt\mathrm { V } = 189 \sin 120 \pi \mathrm { t }V=189sin120πt

Solve the Problem $\mathrm { V } = 189 \sin 120 \pi \mathrm { t }$