The voltage v(t) 5 359.3 cos(vt) volts is applied to a load consisting of a 10- resistor in parallel with a capacitive reactance XC 5 25 Ω.

Question

The voltage v(t) 5 359.3 cos(vt) volts is applied to a load consisting of a 10- resistor in parallel with a capacitive reactance XC 5 25 Ω. Calculate (a) the instantaneous power absorbed by the resistor, (b) the instantaneous power absorbed by the capacitor, (c) the real power absorbed by the resistor, (d) the reactive power delivered by the capacitor, and (e) the load power factor.

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Nho 4 years 2021-07-15T22:42:44+00:00 1 Answers 98 views 0

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  1. giabao
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    2021-07-15T22:43:56+00:00
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    Answer:

    (a). The instantaneous power absorbed by the resistor is 6454.82(1+\cos2\omega t)

    (b). The instantaneous power absorbed by the capacitor is 2581.92\cos(2\omega t+90)

    (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 v(t)= 359.3\cos(\omega t)

    Resistor = 10 Ω

    Capacitive reactance = 25 Ω

    We need to calculate the current by the resistance

    Using formula of current

    I_{1}=\dfrac{V}{R}

    Put the value into the formula

    I_{1}=\dfrac{359.3\cos(\omega t)}{10}

    I_{1}=35.93\cos(\omega t)

    We need to calculate the current by the capacitance

    Using formula of current

    I_{2}=\dfrac{V}{C}

    Put the value into the formula

    I_{2}=\dfrac{359.3\cos(\omega t)}{25}

    I_{2}=14.3\cos(\omega t)

    (a). We need to calculate the instantaneous power absorbed by the resistor

    Using formula of power

    P_{1}=v\times I_{1}

    Put the value into the formula

    P_{1}=359.3\cos(\omega t)\times35.93\cos(\omega t)

    P_{1}=12909.649\cos^2(\omega t)

    P_{1}=\dfrac{12909.649}{2}\times2\cos^2(\omega t)

    P_{1}= 6454.82(1+\cos2\omega t)….(I)

    (b). We need to calculate the instantaneous power absorbed by the capacitor

    Using formula of power

    P_{2}=v\times I_{2}

    P_{2}=359.3\cos(\omega t)\times14.372\cos(\omega t+90)

    P_{2}=\dfrac{5163.85}{2}(\cos(\omega t-\omega t+90)+(\cos2\omega t+90)

    P_{2}=2581.92\cos(2\omega t+90)….(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

    Load\ factor =\dfrac{real\ power\absorbed\ by\ resistance}{real\ power\absorbed\ by\ capacitance}    

    Put the value into the formula

    load\ factor =\dfrac{6454.82}{2581.92}

    load\ factor = 2.5

    Hence, (a). The instantaneous power absorbed by the resistor is 6454.82(1+\cos2\omega t)

    (b). The instantaneous power absorbed by the capacitor is 2581.92\cos(2\omega t+90)

    (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

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