Question

According to Equation 20.7, an ac voltage V is given as a function of time t by V = Vo sin 2ft, where Vo is the peak voltage and f is the frequency (in hertz). For a frequency of 48.3 Hz, what is the smallest value of the time at which the voltage equals one-half of the peak-value?

Answer 1

`1.74* 10^(-3)\ s`

Step-by-step explanation:

Given:

The equation for voltage (V) in terms of frequency (f) and time (t) is given as:

`V=V_0 sin(2\pi ft)`

Where,
`V_0` is the peak voltage.

Frequency (f) = 48.3 Hz

Voltage (V) = Half of peak voltage
`(V_0)`=
`0.5V_0`

Now, plug in the given values in the above equation and solve for 't'. This gives,

`0.5V_0=V_0* \sin(2*\pi* 48.3* t)\\\\\sin(96.6\pi t)=0.5\\\\96.6\pi t=\sin^(-1)(0.5)\\\\96.6\pi t=(\pi)/(6)\\\\t=(1)/(96.6* 6)=1.74* 10^(-3)\ s`

Therefore, the smallest value of the time at which the voltage equals one-half of the peak-value is
`1.74* 10^(-3)\ s`