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2 years ago

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An RLRL circuit with with a 1212-ΩΩ resistor and a 0.06−H0.06−H inductor carries a current of 1A1A at t=0t=0, at which time a vo

ltage source E(t)=12cos(120t)E(t)=12cos⁡(120t) VV is added. Determine the subsequent inductor current I(t)I(t). Hint: use the equation LdIdt+RI=E(t)LdIdt+RI=E(t) where L=L=inductance in henry R=R=resistance in ohm I=I=current in amp E=E=voltage in volts

1 answer:

tankabanditka [31]2 years ago

4 0

Answer:

The equation of current is $I=0.86 cos \left (120 t + \frac{\pi}{2} \right )$

Explanation:

Resistance, R = 12 ohm

Inductance, L = 0.06 H

E (t) = 12 cos (120 t)

Compare with the standard equation,

$E=E_{0}cos (2\pi ft)$

$2\pi ft = 120 t \\\\\\w = 2\pi f = 120 rad/s$

So, the inductive reactance is

XL = w L = 120 x 0.06 = 7.2 ohm

The impedance of the circuit is

$Z =\sqrt{12^2+7.2^2}\\\\Z = 14 ohm$

The current leads by 90degree so the equation of current is

$I=\frac{Eo}{Z} cos \left (120 t + \frac{\pi}{2} \right )\\\\I=\frac{12}{14} cos \left (120 t + \frac{\pi}{2} \right )\\\\I=0.86 cos \left (120 t + \frac{\pi}{2} \right )$

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