What is the inductive time constant of the circuit?

+3 votes

Consider the circuit in the figure below, taking emf = 6.00 V, L = 6.00 mH, and R = 7.40 Ω.

image
(a) What is the inductive time constant of the circuit?
 ms

(b) Calculate the current in the circuit 250 µs after the switch is closed.
 A

(c) What is the value of the final steady-state current?
 A

(d) After what time interval does the current reach 80.0% of its maximum value?
 ms
asked Mar 6, 2012 in Physics by apush ~Expert~ (552 points)
    

1 Answer

+2 votes

Part A:

Inductive time constant = L / R  (*where L=inductance, R=resistance)

Part B:

If a resistor and indcutor are connected in a series to a battery of emf  e   at time t=0, the current in the circuit varies intime according to the following formula:

I= e/R (1-e^(-t/t)  (*where e=emf, R=resistance, t=time, t=inductive time constant)

plugging in the corresponding values will give you I.

 

Part C:

The final steady state current, is defined by Ohm's Law, as:

I = e/R

 

Part D:

The current reaches 80% of its max value when:

0.8 = (1-e^(-t/t)  

solving for t, we get:

0.8-1= -e^(-t/t)  

0.2 = e^(-t/t) 

so,

t =  - ln (0.2) * t

answered Mar 13, 2012 by Joey33 ~Expert~ (1,216 points)
edited Mar 13, 2012 by Joey33

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