Solve the problem.
-The current I, in amperes, flowing through a particular ac (alternating current) circuit at time t seconds is $$I = 220 \sin ( 65 \pi t )$$ What is the period and amplitude of the current?
A) period $= \frac { \pi } { 120 }$ second, amplitude $= 60$
B) period $= \frac { 1 } { 30 }$ second, amplitude $= 120$
C) period $= 60 \pi$ seconds, amplitude $= \frac { 1 } { 30 }$
D) period $= \frac { 1 } { 480 }$ second, amplitude $= 480$

Question

Solve the problem.
-The current I, in amperes, flowing through a particular ac (alternating current) circuit at time t seconds is $$I = 220 \sin ( 65 \pi t )$$ What is the period and amplitude of the current? 
A) period $= \frac { \pi } { 120 }$ second, amplitude $= 60$
B) period $= \frac { 1 } { 30 }$ second, amplitude $= 120$
C) period $= 60 \pi$ seconds, amplitude $= \frac { 1 } { 30 }$
D) period $= \frac { 1 } { 480 }$ second, amplitude $= 480$

Quizplus · Accepted Answer

The amplitude of the current is the coefficient of the sine function, which is 220. The period of the sine function is given by $ \frac{2\pi}{\omega} $, where $ \omega $ is the angular frequency. Here, $ \omega = 65\pi $, so the period is $ \frac{2\pi}{65\pi} = \frac{1}{32.5} $, which simplifies to $ \frac{2}{65} $ or approximately $ \frac{1}{32.5} $, closest to $ \frac{1}{30} $ second. Therefore, the correct answer is period $= \frac{1}{30}$ second, amplitude $= 220$, but since 220 is not an option and there seems to be a mistake in the amplitude options provided, the closest match by period is option B, acknowledging a discrepancy in the amplitude values provided in the question and choices.

Solve the Problem I=220sin⁡(65πt)I = 220 \sin ( 65 \pi t )I=220sin(65πt)

Solve the Problem $I = 220 \sin ( 65 \pi t )$