Home/Class 12/Physics/

## QuestionPhysicsClass 12

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.

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 ) $${t}={100}{m}{s}={0.1}{s}$$
$${Q}={0.5}{\left[{0.1}+{0.1}\times{e}-^{{{0.1}/{0.01}}}-{0.01}\right]}$$
$$={0.045}{C}={45}{m}{c}.$$