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An inductor coil of resistance \({10}\Omega\) and inductance 120 mH is connected across a battery of emf 6 V and internal resistance \({2}\Omega\). Find the charge which flows through the inductor in (a) 10 ms, (b) 20ms and (c ) 100ms after the connections are made.

Answer

Here \({L}={120}{m}{H}={0.120}{H}\)
\({R}={10}{W},{E}{m}{f}={6},{r}={2}\)
\({i}={i}_{{0}}{\left({1}-{e}^{{{1}-{e}^{{-{t}}}/{\left(\tau\right)}}}\right.}\)
Now \({d}{Q}={i}{\left.{d}{t}\right.}={i}_{{0}}{\left({1}-{e}^{{-{t}/\tau}}\right)}{\left.{d}{t}\right.}\)
\({Q}=\int{d}{Q}\)
\(={\int_{{0}}^{{{t}}}}{i}_{{{0}}}={\int_{{{0}}}^{{{t}}}}{i}_{{{0}}}{\left({1}-{e}^{{-{t}/\tau}}\right)}{\left.{d}{t}\right.}\)
\({i}_{{0}}{\left[{\int_{{{0}}}^{{{t}}}}{\left({e}^{{-{t}/\tau}}\right)}{\left.{d}{t}\right.}\right]}\)
\({i}_{{0}}{\left[{t}-{\left(-{\left(\tau\right)}\right)}{\int_{{{0}}}^{{{1}}}}{\left({e}^{{-{t}/\tau}}\right)}{\left({\left.{d}{t}\right.}\right)}\right]}\)
\({i}_{{0}}{\left[{t}+{t}{\left\lbrace{e}^{{-{t}/\tau}}\right\rbrace}-{t}\right]}\)
\(={i}_{{0}}{\left({1}-\right.}\)tau(e))[(1-(tau)e)int_(0)^(1)(e^(-t//tau))(dt)]\({N}{o}{w},\)i_(0)=(6)/(10+2)=(6/12)=0.5A.\(\)(tau)=(L/R)=(120/12)=0.01\({\left({a}\right)}\)t=0.01s\({s}{o},\)Q=0.5[0.01xx+0.01e^(-0.1//0.01)-0.01]\(={0.00108}={1.8}\times{10}^{{-{3}}}={1.8}{m}{Q}\)
(b) \({t}={20}{m}{s}={2}\times{10}^{{-{2}}}{s}={0.02}{s}\)
\({Q}={0.5}{\left[{0.02}+{0.01}\times{e}^{{-{0.02}/{0.01}}}\right.}\)
\(-{\left({0.01}\right)}{]}\)
\(={0.005676}={5.6}\times{10}^{{-{3}}}{C}\)
\(={5.6}{m}{C}\)
(c ) \({t}={100}{m}{s}={0.1}{s}\)
\({Q}={0.5}{\left[{0.1}+{0.1}\times{e}-^{{{0.1}/{0.01}}}-{0.01}\right]}\)
\(={0.045}{C}={45}{m}{c}.\)
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