A coil of self inductance 10 mH and resistance

Question:

A coil of self inductance $10 \mathrm{mH}$ and resistance $0.1 /$ is connected through a switch to a battery of internal resistance $0.9$. After the switch is closed, the time taken for the current to attain $80 \%$ of the saturation value is

$[$ take $\ln 5=1.6]$

  1. (1) $0.324 \mathrm{~s}$

  2. (2) $0.103 \mathrm{~s}$

  3. (3) $0.002 \mathrm{~s}$

  4. (4) $0.016 \mathrm{~s}$


Correct Option: , 4

Solution:

(4) $I=I_{0}\left(1-e^{-\frac{R t}{L}}\right)$          Here $\mathrm{R}=\mathrm{R}_{\mathrm{L}}+\mathrm{r}=1 \Omega$

$0.8 I_{0}=I_{0}\left(1-e^{\frac{t}{.01}}\right)$

$\Rightarrow 0.8=1-e^{-100 t}$

$\Rightarrow e^{-100 t}=0.2=\left(\frac{1}{5}\right)$

$\Rightarrow 100 \mathrm{t}=\ln 5 \Rightarrow \mathrm{t}=\frac{1}{100} \ln 5=0.016 \mathrm{~s}$

 

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