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QJEEM20Ph63d9: JEE Main 2020 Physics - Class 12 JEEM20Ph63d9 Solution: A shown in the figure, a battery of emf is connected to an…

A shown in the figure, a battery of emf ε\varepsilon 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 : JEE Main 2020 (Online) 8th January Evening Slot Physics - Electromagnetic Induction Question 88 English

Multiple Choice Options

  • A εLeR2{{\varepsilon L} \over {e{R^2}}}
  • B εLR2{{\varepsilon L} \over {{R^2}}}
  • C εLR2(1−1e){{\varepsilon L} \over {{R^2}}}\left( {1 - {1 \over e}} \right)
  • D εReL2{{\varepsilon R} \over {e{L^2}}}

Step-by-Step Verified Solution

i=i0(1−e−t/tc)i = {i_0}\left( {1 - {e^{ - t/{t_c}}}} \right)

q = ∫0tcidt\int\limits_0^{{t_c}} {idt}

= ∫0tc∈R(1−e−t/tc)dt\int\limits_0^{{t_c}} {{ \in \over R}\left( {1 - {e^{ - t/{t_c}}}} \right)dt}

= ∈R[t−e−t/tc(−1tc)]0tc{ \in \over R}\left[ {t - {{{e^{ - t/{t_c}}}} \over {\left( { - {1 \over {{t_c}}}} \right)}}} \right]_0^{{t_c}}

= ∈R[(tc+tce−1)−(0+tc)]{ \in \over R}\left[ {\left( {{t_c} + {t_c}{e^{ - 1}}} \right) - \left( {0 + {t_c}} \right)} \right]

= ∈Rtc.1e{ \in \over R}{t_c}.{1 \over e}

= ∈R.LR.1e{ \in \over R}.{L \over R}.{1 \over e}

= ∈LeR2{{ \in L} \over {e{R^2}}}

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