Question

A coil of inductance $8.4mH$ and resistance $6\Omega$ is connected to $12V$ battery in series. The current in the coil is $1.0A$ at approximately the time

• A. $500s$
• B. $20s$
• C. $35ms$
• D. $1ms$

Answer 1

The correct option is D $1ms$

$L=8.4\times {10}^{-3}HR=6\Omega V=12vi=1At=?$

From current-time equation in $R-L$ circuit,

$i=i_0(1-e^{-\frac{t}{T_L}})$

here, $T_L=\frac{L}{R}$

or $T_L=\frac{8.4\times {10}^{-3}}{6}=1.4\times {10}^{-3}s$

and $i=\frac{V}{R}=\frac{12}{6}=2A$

$i=given=1A$

$\therefore 1=2(1-e^{\frac{-t}{1.4\times {10}^{-3}}})\frac{1}{2}=1-\frac{1}{e^{(t\times \frac{{10}^3}{1.4})}}$

$e^{(t\times \frac{{10}^3}{1.4})}=2$ , take log both side

$\ln 2=t\times \frac{{10}^3}{1.4},t=(1.4\times \ln 2)\times {10}^{3}sect=0.9704ms\approx 1ms$