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5 March, 10:48

A series RLC circuit containing a resistance of 12Ω, an inductance of 0.15H and a capacitor of 100uF are connected in series across a 100V, 50Hz supply. Calculate the total circuit impedance, the circuit's current, and the power factor.

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  1. 5 March, 14:26
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    Impedance = 19.44ohms

    Current = 5.14A

    Power factor = 0.62

    Explanation:

    Impedance in an RLC AC circuit is defined as the total opposition to the flow of current in the resistor, inductor and capacitor.

    Impedance Z = √R² + (Xl-Xc) ²

    Where R is the resistance = 12Ω

    Inductance L = 0.15H

    Capacitance C = 100uF = 100*10^-6F

    Since Xl = 2πfL and Xc = 1/2πfC where f is the frequency.

    Xl = 2π*50*0.15

    Xl = 15πΩ

    Xl = 47.12Ω

    Xc = 1/2π*50*100*10^-6

    Xc = 100/π Ω

    Xc = 31.83Ω

    Z = √12² + (47.12-31.83) ²

    Z = √144+233.78

    Z = 19.44Ω

    Impedance = 19.44ohms

    To calculate the circuit current, we will use the expression V=IZ where V is the supply voltage = 100V

    I = V/Z = 100/19.44

    I = 5.14Amperes

    To calculate the power factor,

    Power factor = cos (theta) where;

    theta = arctan (Xl-Xc) / R

    theta = arctan (47.12-31.83) / 12

    theta = arctan (15.29/12)

    theta = arctan1.27

    theta = 51.78°

    Power factor = cos51.78°

    Power factor = 0.62
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