Answer:
[tex]P(3600)=593.247W[/tex]
Explanation:
First, let's find the voltage through the resistor using ohm's law:
[tex]V=IR=20*8=160V[/tex]
AC power as function of time can be calculated as:
[tex]P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)[/tex] (1)
Where:
[tex]\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency[/tex]
Because of the problem doesn't give us additional information, let's assume:
[tex]\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi[/tex]
Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):
[tex]P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W[/tex]
The amount of energy that is dissipated by the resistor =4608kJ
Energy dissipated = I²Rt
Where current ( I ) = 8.0A
Resistor (R) = 20-Ω
Time (t) = 1 h
To convert 1 hour to second times by 3600
This is because, 60 seconds make 1 minute and 60 minutes make 1 hour.
Therefore energy dissipation= 8² × 20 × 3600
= 64 × 20 × 3600
= 4608000j
= 4608kJ
Learn more about energy dissipation here:
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