Solve the problem.
-Two voltages, $V _ { 1 } = 70 \sin ( 100 \pi t )$ and $V _ { 2 } = 240 \cos ( 100 \pi t )$, are added in a circuit to produce the sum $V = V _ { 1 } + V _ { 2 }$. Graph $V$ in $[ - 0.02,0.02,0.002 ]$ by $[ - 250,250,20 ]$ and use the graph to estimate values for $\mathrm { a }$ and $\mathrm { k }$ so that $\mathrm { V } = \mathrm { a } \sin ( 100 \pi \mathrm { t } + \mathrm { k } )$. Round $\mathrm { k }$ to the nearest ten-thousandth, if necessary.
A) $\mathrm { a } = 25 ; \mathrm { k } = 1.287$
B) $\mathrm { a } = 250 ; \mathrm { k } = 404.3233$
C) $a = 250 ; k = 1.287$
D) $\mathrm { a } = 250 ; \mathrm { k } = 89.1565$

Question

Quizplus · Accepted Answer

To find the amplitude $a$ and phase shift $k$ for the sum of the two voltages, we can use trigonometric identities to combine them into a single sinusoidal function. The given voltages are $V_1 = 70\sin(100\pi t)$ and $V_2 = 240\cos(100\pi t)$. We can convert $V_2$ into a sine function using the identity $\cos(x) = \sin(x + \frac{\pi}{2})$, making $V_2 = 240\sin(100\pi t + \frac{\pi}{2})$. Combining $V_1$ and $V_2$ and using trigonometric identities, we find that the amplitude $a$ is the square root of the sum of the squares of the coefficients of the sine and cosine terms (in this case, the same since we converted everything into sine), and $k$ is the phase shift calculated from the arctan of the ratio of the coefficients. This results in $a = \sqrt{70^2 + 240^2} = 250$ and $k$ calculated from the relationship between the coefficients, leading to an estimated phase shift that matches the option given.

Solve the Problem V1=70sin⁡(100πt)V _ { 1 } = 70 \sin ( 100 \pi t )V1​=70sin(100πt)

Solve the Problem $V _ { 1 } = 70 \sin ( 100 \pi t )$