Solve.
-In the study of alternating electric current, capacitive voltage is given by $V _ { C } = - \frac { i } { 2 \pi f } \cos 2 \pi f \mathrm { ft }$, where $\mathrm { f }$ is the number of cycles per second, $\mathrm { i }$ is the maximum current, and $\mathrm { t }$ is the time in seconds. Solve the equation for $\mathrm { t }$.
A) $\mathrm { t } = \frac { 1 } { 2 \pi \mathrm { f } } \arccos \left( \frac { 2 \pi \mathrm { fV } _ { \mathrm { C } } } { \mathrm { i } } \right)$
B) $t = 2 \pi f \arccos \left( - \frac { 2 \pi f V _ { C } } { i } \right)$
C) $\mathrm { t } = \frac { 1 } { 2 \pi \mathrm { f } } \arccos \left( - \frac { 2 \pi \mathrm { fV } _ { \mathrm { C } } } { \mathrm { i } } \right)$
D) $t = 2 \pi f \cos \left( - \frac { 2 \pi \mathrm { fV } _ { \mathrm { C } } } { \mathrm { i } } \right)$

Question

Quizplus · Accepted Answer

To solve for $t$, we isolate $t$ from the given equation $V _ { C } = - \frac { i } { 2 \pi f } \cos 2 \pi f \mathrm { ft }$. Rearranging the equation to solve for $\cos 2 \pi f \mathrm { ft }$ and then for $t$, we get $t = \frac { 1 } { 2 \pi \mathrm { f } } \arccos \left( - \frac { 2 \pi \mathrm { fV } _ { \mathrm { C } } } { \mathrm { i } } \right)$.

Solve -In the Study of Alternating Electric Current, Capacitive Voltage VC=−i2πfcos⁡2πfftV _ { C } = - \frac { i } { 2 \pi f } \cos 2 \pi f \mathrm { ft }VC​=−2πfi​cos2πfft

Solve
-In the Study of Alternating Electric Current, Capacitive Voltage $V _ { C } = - \frac { i } { 2 \pi f } \cos 2 \pi f \mathrm { ft }$