Question

A resistor, an ideal capacitor, and an ideal inductor are connected in parallel to a source of an alternating voltage of 160 V at a frequency of 250 Hz. A current of 2 A flows through the resistor and a current of 0.8 A flows through the inductor. The total current through the circuit is 2.5 A. Assess the resistance of the resistor, the capacity of the ideal capacitor, and the inductance of the ideal inductor (presume that IC > IL).

Answer 1

Answer:

R = 46,25 (Ω)

L = 0,07363 (H)

C = 2,7 *10⁻⁶ (F)

Explanation:

Statement problem does not specify if  160 (V) is peak voltage or RMS value. If 160 (V) is the peak value then for a sinusoidal wave,  RMS value is  V(rms) = 160 /√3

V(rms) =  92,49 (V)

In each branch: we have

V = I*Z         V = voltage through the impedance in (V) , I current in (A)

and Z impedance in Ω

Resistor case    Z = R  then V = 92,49 = 2 * R

R = 92,49/2       R = 46,25 (Ω)

Inductor case   |Z| = wL    Then Z = 2*π*f*L   V = 0,8 * |wl|

Inductor L in (H)       |wL| * 0,8 = 92,49   |wL| = 92,5/0,8    w = 2*π*f   2*3,14*250= 1570

1570*L = 92,49/0,8

L = 0,07363 (H)

In the case of the capacitor

|Z|  = 1/wc    = 1/1570*c

The current is 2 + 0,8 = 2,8          2,8 - 2,5 = 0,3 (A)

Again  V = I*Z        92,49 = 0,3 /1570*C

C = 0,3 / 1570*92,49

C = 2,7 *10⁻⁶ (F)