Question

An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 Ω) are connected in series. A 44.0-Hz AC generator connected in series to these elements produces a maximum current of 385 mA in the circuit.
(a) Calculate the required maximum voltage ΔVmax.
(b) Determine the phase angle by which the current leads or lags the applied voltage.

Answer 1

Answer:

(A) Maximum voltage will be equal to 333.194 volt

(B) Current will be leading by an angle 54.70

Explanation:

We have given maximum current in the circuit $i_m=385mA=385\times 10^{-3}A=0.385A$

Inductance of the inductor $L=400mH=400\times 10^{-3}h=0.4H$

Capacitance $C=4.43\mu F=4.43\times 10^{-3}F$

Frequency is given f = 44 Hz

Resistance R = 500 ohm

Inductive reactance will be $x_l=\omega L=2\times 3.14\times 44\times 0.4=110.528ohm$

Capacitive reactance will be equal to $X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 44\times 4.43\times 10^{-6}}=816.82ohm$

Impedance of the circuit will be $Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{500^2+(816.92-110.52)^2}=865.44ohm$

So maximum voltage will be $\Delta V_{max}=0.385\times 865.44=333.194volt$

(B) Phase difference will be given as $\Phi =tan^{-1}\frac{X_C-X_L}{R}=\frac{816.92-110.52}{500}=54.70$

So current will be leading by an angle 54.70