4 years ago

I'm really bad at measurements so I don't understand this.

Engineering

1 answer:

cricket20 [7]4 years ago

6 0

Answer:

the answer should be 16

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An AC circuit has a resistor, capacitor and inductor in series with a 120 V, 60 Hz voltage source. The resistance of the resisto

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(i) 3.5385 ohm, 3.768 ohm (ii) 39.89 A (III) 4773.857 W (vi) 348 var (vii) 0.9973 (viii) 4.1796°

Explanation:

We have given voltage V =120 volt

Frequency f=60 Hz

Resistance R =3 ohm

Inductance L =0.01 H

Capacitance C =0.00075 farad

(i) reactance of of inductor $X_L=\omega L=2\pi fL=2\times 3.14\times 60\times 0.01=3.768ohm$

$X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 60\times 0.00075}=3.5385ohm$

(ii) Total impedance $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{3^2+(3.768-3.5385)^2}=3.008ohm$

Current $i=\frac{V}{Z}=\frac{120}{3.008}=39.89A$

(viii) power factor $cos\Phi =\frac{R}{Z}=\frac{3}{3.008}=0.9973$

(VII) $cos\Phi =0.9973$

$\Phi =4.1796^{\circ}$

So power factor angle is 4.1796°

(iii) Apparent power $P=VICOS\Phi =120\times 39.89\times 0.9973=4773.875W$

(vi) Reactive power $Q=VISIN\Phi =120\times 39.89\times SIN4.17^{\circ}=348var$

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