Respuesta :
Answer:
(a). The instantaneous power absorbed by the resistor is [tex]6454.82(1+\cos2\omega t)[/tex]
(b). The instantaneous power absorbed by the capacitor is [tex]2581.92\cos(2\omega t+90)[/tex]
(c). The real power absorbed by the resistor is 6454.82 watts.
(d). The real power absorbed by the capacitor is 2581.92 watts.
(e) The load factor is 2.5
Explanation:
Given that,
Voltage [tex]v(t)= 359.3\cos(\omega t)[/tex]
Resistor = 10 Ω
Capacitive reactance = 25 Ω
We need to calculate the current by the resistance
Using formula of current
[tex]I_{1}=\dfrac{V}{R}[/tex]
Put the value into the formula
[tex]I_{1}=\dfrac{359.3\cos(\omega t)}{10}[/tex]
[tex]I_{1}=35.93\cos(\omega t)[/tex]
We need to calculate the current by the capacitance
Using formula of current
[tex]I_{2}=\dfrac{V}{C}[/tex]
Put the value into the formula
[tex]I_{2}=\dfrac{359.3\cos(\omega t)}{25}[/tex]
[tex]I_{2}=14.3\cos(\omega t)[/tex]
(a). We need to calculate the instantaneous power absorbed by the resistor
Using formula of power
[tex]P_{1}=v\times I_{1}[/tex]
Put the value into the formula
[tex]P_{1}=359.3\cos(\omega t)\times35.93\cos(\omega t)[/tex]
[tex]P_{1}=12909.649\cos^2(\omega t)[/tex]
[tex]P_{1}=\dfrac{12909.649}{2}\times2\cos^2(\omega t)[/tex]
[tex]P_{1}= 6454.82(1+\cos2\omega t)[/tex]....(I)
(b). We need to calculate the instantaneous power absorbed by the capacitor
Using formula of power
[tex]P_{2}=v\times I_{2}[/tex]
[tex]P_{2}=359.3\cos(\omega t)\times14.372\cos(\omega t+90)[/tex]
[tex]P_{2}=\dfrac{5163.85}{2}(\cos(\omega t-\omega t+90)+(\cos2\omega t+90)[/tex]
[tex]P_{2}=2581.92\cos(2\omega t+90)[/tex]....(II)
(c). We need to find the real power absorbed by the resistor
Using equation (I)
The real power absorbed by the resistor is 6454.82 watts.
(d). We need to find the real power absorbed by the capacitor
Using equation (II)
The real power absorbed by the capacitor is 2581.92 watts.
(e). We need to calculate the load factor
Using formula of load factor
[tex]Load\ factor =\dfrac{real\ power\absorbed\ by\ resistance}{real\ power\absorbed\ by\ capacitance}[/tex]
Put the value into the formula
[tex]load\ factor =\dfrac{6454.82}{2581.92}[/tex]
[tex]load\ factor = 2.5[/tex]
Hence, (a). The instantaneous power absorbed by the resistor is [tex]6454.82(1+\cos2\omega t)[/tex]
(b). The instantaneous power absorbed by the capacitor is [tex]2581.92\cos(2\omega t+90)[/tex]
(c). The real power absorbed by the resistor is 6454.82 watts.
(d). The real power absorbed by the capacitor is 2581.92 watts.
(e) The load factor is 2.5
