Question

The time constant of L-R circuit is:
a. LR
b. $${\displaystyle \frac{L}{R}}$$
c. $${\displaystyle \frac{R}{L}}$$
d. $${\displaystyle \frac{1}{LR}}$$

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 $$36.8\mathrm{\%}$$ and $63.2\mathrm{\%}$ 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 ${\displaystyle \frac{1}{e}}$ times and the current to become $\left(1-{\displaystyle \frac{1}{e}}\right)$ times of their maximum achievable values.
Voltage across inductor is given as:
$V_L(t)=V_0{e}^{{\displaystyle \frac{-Rt}{L}}}$ ----(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:
$I(t)=I_0(1-e^{{\displaystyle \frac{-Rt}{L}}})$----(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:
$V_0{e}^{{\displaystyle \frac{-R\tau }{L}}}={\displaystyle \frac{V_0}{e}}$
$\Rightarrow {\displaystyle \frac{-R\tau }{L}}=-1$
$\Rightarrow \tau ={\displaystyle \frac{L}{R}}$

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

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

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 $I_0={\displaystyle \frac{V_0}{R}}$.