a time consant and relate it RL circuit P 00LEM A 12.6-V battery is in circuit w

Question

a time consant and relate it RL circuit P 00LEM A 12.6-V battery is in circuit with 30.0-mw nductor and a 0.150-? resisto. The swech is closed at t-o. (a) Find the time constant of the circuit. (b) Find the ourrent after one time constant has elapsed. (e) Find the voitage drops across the resistor whenO and one tim constant. (d) what's the rate of change of the current after time constant? STRATEGY Part (a) requires only subsitution into the of time constant. With this current afber one time constant can be found, and multiplying thid current by the resistance yields the voltage drep acress the resistor after one time constant. wgh the votage drop and Kirchhof's loop law, can be substituted ito che proper to obtain the rate of change of the current (A) What's the time constant of the circut? () Find the ourrent after one time constant has elapsed (C) Find the voltage dreps acress the resistance whentO e one time constant (D) What's the rate of change of the current after one time REMARKS The values used in this problem were taken from actual components salvaged from the starter system of a car inducters are sometimes referred to as "chokes" because they temporarily chelke off the current. n soving part (d) we traversed the orcuit in the direction of positive current, so the voitage dfference acnoss the battery was postive and the fferences across the resistor and inductor were negative QUESTION Mter two time constants of time have elapsed the current in the circuit how many tirmes the firal ament, A12.6-V batery is in series with a resistance of 0.354 ? and (a) After a long time, what is the current in the circut? (b) what is the ourrent after one time constam? (c) what's voltage drop across the inductor this (d) Find the inductance if the time cositant is o.334

Explanation / Answer

Imax = V*R

Imax = 12.6*0.354

Imax = 4.4604 amps

so current after one time constant,

I = Imax ( 1- e-1)

I = Imax*0.632

I = 2.8195 amps

Voltage drop across inductor after this one time constant

VL = -(2.4 - 2.8195*0.354)

VL = - 1.402 volts

L = time constant * resistance

L = 0.134 * 0.354

L = 0.047436 H or 47 mH

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