Question

The current L-R circuit decreases to 1/3 part of its max. Steady value in 5 sec. Calculate the time constant of the circuit.​

Answer 1

Given:

Current decreases to 1/3rd of its steady value in time = 5 sec

To find:

Time constant of the circuit.

Solution:

As we know that at any time =t

i = i₀(1- e^(-t/τ))

τ is the time constant.

t is the time taken.

i and i₀ are final and initial current respectively.

Since the current decreases to 1/3rd of its initial value:

Therefore, 1/3 i₀ =  i₀ (1- e^(-5/τ))

1/3 =  (1- e^(-5/τ))

1- 1/3 =  e^(-5/τ))

2/3 =  e^(-5/τ))

ln(2/3) = -5/τ

-0.405 = -5/τ

τ= 5/ 0.405

τ = 12.346

Therefore the time constant of the circuit will be 12.346.