Question

GOAL Calculate a time constant and relate it to current in an RL circuit.

PROBLEM A 12.6-V battery is in a circuit with a 30.0-mH inductor and a 0.150-? resistor. The switch is closed at t = 0. (a) Find the time constant of the circuit. (b) Find the current after one time constant has elapsed. (c) Find the voltage drops across the resistor when t = 0 and t = one time constant. (d) What's the rate of change of the current after one time constant?

STRATEGY Part (a) requires only substitution into the definition of time constant. With this value and Ohm's law, the current after one time constant can be found, and multiplying this current by the resistance yields the voltage drop across the resistor after one time constant. With the voltage drop and Kirchhoff's loop law, the voltage across the inductor can be found. This value can be substituted into the proper equation to obtain the rate of change of the current.

SOLUTION (A) What's the time constant of the circuit? L = 3.00 x 10 Substitute the inductance L and resistance R into the equation, finding the time constant. 0.200 s (B) Find the current after one time constant has elapsed E 12.6 V First, use Ohm's law to compute the final value of the current after many time constants have elapsed. max R 0.150 84.0 A 11r = (0.632)Imax = (0.632)(84.0 A) = 53.1 A After one time constant, the current rises to 63.2% of its final value. (C) Find the voltage drops across the resistance when t 0 and t one time constant. Initially, the current in the circuit is zero, so, from Ohm's law, the voltage across the resistor is zero AVR(t-0 s)-(0 A)(0.150 (1) o Next, using Ohm's law, find the magnitude of the voltage drop across the resistor after one time constant. AVR(t-0.200 s)-(53.1 A)(0.150 ?)-7.97 V (D) What's the rate of change of the current after one time constant? Using Kirchhoff's voltage rule, calculate the voltage drop across the inductor at that time. Solve for AV AVL =-E-AVR =-12.6 V-(-7.97 V)--4.6 V Now solve for ?1At and substitute. ?? -4.6V 3.00 x 10-2H150 AVs

Answer 1

Epsilon_L = -L delta I/delta T using ohm's law epsilon = IR material implecation IR = -L delta I/delta T material implecation integrate dI/I = integrate -R/L delta t = ln I = -Rt/L + C at t = 0, I = I_0 C = ln I_0 = ln I - ln I_0 = -Rt/L Time constant , c = L/R = I = I_0 e^-t/c t = 2c material implecation I = I_0 e^-2 or I = 0.1354I_0 so current in inductor will be 0, 135 times initial inductance = L Epsilon = 12.6v R = 0.386 ohms (a) Inductor at steady state behaves as zero resistance so circuit will be theta_sigma = epsilon/R = 12.6/0.386 amp as I = I_0(1-e^--t/z) z = time constant \r\n (b) so after t = z I(t = z) = I_0(1-e^-1) material implecation I(t = z) = 20.63 amp \r\n delta v_L = -IR + epsilon or deltav_l = 12.6 - (20.63)(0.386) material implecation delta v_l = 4.65 volt \r\n (d) given tau = 0.129sec as tau = L/R material implecation (0.129) = L/(0.386) material implecation L = 0.05 henry