Answer:
[tex]1.74\times 10^{-3}\ s[/tex]
Explanation:
Given:
The equation for voltage (V) in terms of frequency (f) and time (t) is given as:
[tex]V=V_0 sin(2\pi ft)[/tex]
Where, [tex]V_0[/tex] is the peak voltage.
Frequency (f) = 48.3 Hz
Voltage (V) = Half of peak voltage [tex](V_0)[/tex]= [tex]0.5V_0[/tex]
Now, plug in the given values in the above equation and solve for 't'. This gives,
[tex]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[/tex]
Therefore, the smallest value of the time at which the voltage equals one-half of the peak-value is [tex]1.74\times 10^{-3}\ s[/tex]
