Respuesta :
Answer:
Explanation:
Given
Voltage [tex]V=120\ V[/tex]
Current [tex]I=0.737\sin [303t][/tex]
frequency of the alternating current is given by
[tex]f=\frac{\omega }{2\pi }v[/tex]
[tex]f=\frac{303}{2\pi }[/tex]
[tex]f=48.21\ Hz[/tex]
(b)Mean current [tex]I_m=0.737\ A[/tex]
[tex]I_{rms}=\frac{I_m}{\sqrt{2}}[/tex]
[tex]I_{rms}=\frac{0.737}{\sqrt{2}}[/tex]
[tex]I_{rms}=0.521\ A[/tex]
[tex]R=\frac{V_{rms}}{I_{rms}}[/tex]
[tex]R=\frac{120}{0.521}[/tex]
[tex]R=230.2\ \Omega [/tex]
(c)Average Power
[tex]P=V_{rms}I_{rms}[/tex]
[tex]P=62.52\ W[/tex]
Answer:
a) [tex]f=48.2239\ Hz[/tex] is the frequency of the AC.
b) [tex]R=230.265\ \Omega\\[/tex]
c) [tex]P=62.532\ W[/tex]
Explanation:
Given:
- voltage rating of the bulb, [tex]V_{rms}=120\ V[/tex]
- current in the bulb as the function of time, [tex]I=0.737A.\sin(303t)[/tex]
Comparing the expression of the current with the standard form we have:
a)
Angular frequency,
[tex]\omega=303\ rad.s^{-1}[/tex]
[tex]2\pi.f=303[/tex]
[tex]f=48.2239\ Hz[/tex] is the frequency of the AC.
b)
From the expression of current we have the maximum value of current as:
[tex]I_m=0.737\ A[/tex]
Now the RMS current is:
[tex]I_{rms}=\frac{I_m}{\sqrt{2} }[/tex]
[tex]I_{rms}=\frac{0.737}{\sqrt{2} }[/tex]
[tex]I_{rms}=0.5211\ A[/tex]
So, resistance
[tex]R=\frac{V_{rms}}{I_{rms}}[/tex]
[tex]R=\frac{120}{0.5211}[/tex]
[tex]R=230.265\ \Omega\\[/tex]
c)
We know that average power is given as:
[tex]P=V_{rms}.I_{rms}[/tex]
[tex]P=120\times 0.5211[/tex]
[tex]P=62.532\ W[/tex]
