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  • A coil of inductance and resistance <img alt="100 \Omega" src="https://entrancecorner.oncodecog

A coil of inductance 1 \mathrm{H} and resistance 100 \Omega is connected to a battery of 6 \mathrm{~V}. Determine approximately :
(a) The time elapsed before the current acquires half of its steady - state value.
(b) The energy stored in the magnetic field associated with the coil at an instant \mathrm{15 \mathrm{~ms}} after the circuit is switched on. (Given \mathrm{\ln 2=0.693, \mathrm{e}^{-3 / 2}=0.25} )
 

Option: 1

\mathrm{t}=10 \mathrm{~ms} ; \mathrm{U}=2 \mathrm{~mJ}
 


Option: 2

\mathrm{t}=10 \mathrm{~ms} ; \mathrm{U}=1 \mathrm{~mJ}


Option: 3

\mathrm{t}=7 \mathrm{~ms} ; \mathrm{U}=1 \mathrm{~mJ}


Option: 4

\mathrm{t}=7 \mathrm{~ms} ; \mathrm{U}=2 \mathrm{~mJ}


Answers (1)

best_answer

\mathrm{L=1 H }

\mathrm{R=100 \Omega }

\mathrm{V=6 V}
(a) Growth of current in R-L Circuit is represented by equation

\mathrm{i=i_0\left(1-e^{-t / \tau}\right)}

\mathrm{steady\: state\: current =i_0=\frac{V}{R}=6 \times 10^{-2} }

\mathrm{i=\frac{i_0}{2}=i_0\left(1-e^{-t / \tau}\right)}

\mathrm{\frac{1}{2}=e^{-t / \tau}}

\mathrm{\ln (2)=\frac{t}{\tau}=\frac{t}{L / R}}

\mathrm{\ln (2)=\frac{100 t}{1}}

\mathrm{t=\frac{0.693}{100} \cong 7 \mathrm{~ms}}

the time elapsed before the current acquires half of its steady state value

(b) The energy stored in the magnetic field associated with coil at \mathrm{t=15m}

\mathrm{i =i_0\left(1-e^{-t / \tau}\right) }

  \mathrm{=6 \times 10^{-2}\left(1-e^{-15 / 10}\right) }

\mathrm{i=6 \times 10^{-2}\left(\frac{3}{4}\right) =4.5 \times 10^2 \mathrm{~A} }

\mathrm{\text { Energy stored } =\frac{1}{2} Li^2 }

                              \mathrm{=\frac{1}{2} \times 1 \times \frac{81}{4} \times 10^{-4}}

                              \mathrm{=\frac{81}{8} \times 10^{-4} }

                             \mathrm{=10^{-3}J}

Hence 3 is correct option.







 

 

 

 

 

 

 

Posted by

shivangi.bhatnagar

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