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

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

Explanation

Solution

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%36.8\% and 63.2%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 1e\dfrac{1}{e} times and the current to become (1−1e)\left( {1 - \dfrac{1}{e}} \right) times of their maximum achievable values.
Voltage across inductor is given as:
VL(t)=V0e−RtL{V_L}(t) = {V_0}{e^{\dfrac{- Rt}{L}}} ----(1)

Where,
VL{V_L} is voltage across the inductor,
V0{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)=I0(1−e−RtL)I(t) = {I_0}\left( 1 - {e^{\dfrac{- Rt}{L}}}\right)----(2)
Where,
I is current through the circuit,
I0{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=τt = \tau you’ll get:
V0e−RτL=V0eV_0{e^{\dfrac{- R\tau }{L}}} = \dfrac{V_0}{e}
⇒−RτL=−1\Rightarrow { \dfrac{- R\tau }{L}} = - 1
⇒τ=LR\Rightarrow \tau = \dfrac{L}{R}

Step 2
Put this value of τ\tau in eq.(2) to get:
I=I0(1−e−RτL)I=I_0\left ( 1-e^{\dfrac{-R\tau}{L}} \right )
=I0(1−e−RL×LR)= {I_0}\left( {1 - {e^{ - \dfrac{R}{L} \times \dfrac{L}{R}}}} \right)
=I0(1−1e)= {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) LR\dfrac{L}{R}..

Note: Experimentally, it has been noticed that approximately after a time 5τ5\tau the both the voltage across the inductor and current through the circuit saturates. Voltage drop becomes 0 while current reaches to I0=V0R{I_0} = \dfrac{{{V_0}}}{R}.