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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 voltage 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

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Numenius

Answer:

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

Explanation:

Resistance, R = 12 ohm

Inductance, L = 0.06 H

E (t) = 12 cos (120 t)

Compare with the standard equation,

[tex]E=E_{0}cos (2\pi ft)[/tex]

[tex]2\pi ft = 120 t \\\\\\w = 2\pi f = 120 rad/s[/tex]

So, the inductive reactance is

XL = w L = 120 x 0.06 = 7.2 ohm

The impedance of the circuit is

[tex]Z =\sqrt{12^2+7.2^2}\\\\Z = 14 ohm[/tex]

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

[tex]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 )[/tex]

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