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  • For the given L-R network, the time constant of the circuit is equals to <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/08/07/phy-q25-f21.PNG" style="height:143px; width:

For the given L-R network, the time constant of the circuit is equals to

Option: 1

\mathrm{\text { L/R }}


Option: 2

\mathrm{\text { 2L/R }}


Option: 3

\mathrm{\text { L/2R }}


Option: 4

none of these


Answers (1)

best_answer

The impedance of L-R segment is \mathrm{Z=R+j \omega L} If two similar impedences are in parallel,

the equivalent impedance \mathrm{Z_{\text {ef }}=\frac{Z}{2}=\frac{R}{2}+\frac{j \omega L}{2}}

\mathrm{\therefore} The ratio of inductance & resistance for the resulting R – L circuit is equal to

\mathrm{L_{\text {eff }} / R_{\text {eff }}=(L / 2) /(R / 2)=\frac{L}{R}}       \mathrm{\Rightarrow \text { The time constant } \tau=\mathrm{L} / \mathrm{R} \text {. }}

\mathrm{\therefore(A)}

Posted by

Ritika Kankaria

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