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

Answer:

$1.74\times 10^{-3}\ s$

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,

$0.5V_0=V_0\times \sin(2\times\pi\times 48.3\times t)\\\\\sin(96.6\pi t)=0.5\\\\96.6\pi t=\sin^{-1}(0.5)\\\\96.6\pi t=\frac{\pi}{6}\\\\t=\frac{1}{96.6\times 6}=1.74\times 10^{-3}\ s$

Therefore, the smallest value of the time at which the voltage equals one-half of the peak-value is $1.74\times 10^{-3}\ s$