Question

The time constant of L-R circuit is:
a. LR
b. \[\dfrac{L}{R}\]
c. \[\dfrac{R}{L}\]
d. \[\dfrac{1}{{LR}}\]

Answer 1

Hint: Time constant of L-R circuit is a particular amount of tie that can be calculated from the voltage expression across the inductor or from the current expression through the circuit. After a time the same as the time constant voltage and current becomes \[36.8\% \] and $63.2\% $ of their respective maximum values.

Complete answer:
Time constant: Time constant of L-R circuit is the amount of time required for the voltage across L to become $\dfrac{1}{e}$ times and the current to become $\left( {1 - \dfrac{1}{e}} \right)$ times of their maximum achievable values.
Voltage across inductor is given as:
${V_L}(t) = {V_0}{e^{\dfrac{- Rt}{L}}}$ ----(1)

Where,
${V_L}$ is voltage across the inductor,
${V_0}$ is the maximum achievable voltage,
t is time elapsed after switching on the circuit,
R is resistance of the circuit,
L is the inductance of the inductor.

Current through the circuit is given as:
$I(t) = {I_0}\left( 1 - {e^{\dfrac{- Rt}{L}}}\right)$----(2)
Where,
I is current through the circuit,
${I_0}$is the maximum current.

Now in the given question, we have to find the time constant of the L-R circuit.

> Step 1
Use the definition of time constant in eq.(1) to get its value. At $t = \tau $ you’ll get:
$ V_0{e^{\dfrac{- R\tau }{L}}} = \dfrac{V_0}{e}$
$\Rightarrow { \dfrac{- R\tau }{L}} = - 1$
$\Rightarrow \tau = \dfrac{L}{R} $

> Step 2
Put this value of $\tau$ in eq.(2) to get:
$ I=I_0\left ( 1-e^{\dfrac{-R\tau}{L}} \right )$
$ = {I_0}\left( {1 - {e^{ - \dfrac{R}{L} \times \dfrac{L}{R}}}} \right) $
$ = {I_0}\left( {1 - \dfrac{1}{e}} \right) $
Therefore, this verifies that the obtained value of time constant is correct.

Hence, The time constant for the L-R circuit is (b) $\dfrac{L}{R}$..

Note: Experimentally, it has been noticed that approximately after a time $5\tau $ the both the voltage across the inductor and current through the circuit saturates. Voltage drop becomes 0 while current reaches to ${I_0} = \dfrac{{{V_0}}}{R}$.