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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

Question

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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Physics
3 years
2021-07-23T09:57:47+00:00
2021-07-23T09:57:47+00:00 1 Answers
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## Answers ( )

Answer:The equation of current is

Explanation:Resistance, R = 12 ohm

Inductance, L = 0.06 H

E (t) = 12 cos (120 t)

Compare with the standard equation,

So, the inductive reactance is

XL = w L = 120 x 0.06 = 7.2 ohm

The impedance of the circuit is

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