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As shown in the figure, a battery of emf \epsilon is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc (tc is the time constant of the circuit ) is : 

Option: 1 \frac{\epsilon L }{R^{2}} \left ( 1 - \frac{1}{e} \right )
Option: 2 \frac{\epsilon L }{R^{2}}

Option: 3 \frac{\epsilon R }{eL^{2}}

Option: 4 \frac{\epsilon L }{eR^{2}}

Answers (1)






As current at any time t is given as

I = I_{0}\left ( 1 - e^{\frac{-t}{\tau }} \right ) = \frac{\epsilon }{R}\left ( 1 - e^{\frac{-t}{\tau }} \right )

So \mathrm{q}=\int_{0}^{\mathrm{T}_{\mathrm{C}}} \mathrm{idt}


So integrating this will give total charge


q=\frac{\varepsilon}{\mathrm{R}}\left[\mathrm{t}-\frac{\mathrm{e}^{-t / \tau }}{\frac{-1}{ \tau }}\right]_{0}^{ \tau } ; \quad=\frac{\varepsilon}{\mathrm{R}}\left[ \tau + \tau \mathrm{e}^{-1}- \tau \right]


q=\frac{\varepsilon}{\mathrm{R}} \times \frac{1}{\mathrm{e}} \times \frac{\mathrm{L}}{\mathrm{R}} \quad ; \quad q=\frac{\varepsilon \mathrm{L}}{\mathrm{R}^{2} \mathrm{e}}

Hence the correct option is (4).

Posted by

vishal kumar

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