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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2019 · 11 January · Morning Shift · Physics · Alternating Current

In the circuit shown, the switch S_ 1 is closed at time t=0 and the...

In the circuit shown,
In the circuit shown, the switch S_ 1 is closed at time t=0 and the switch S_ 2 is kept open. At some later time (t_ 0
the switch S1S_{1} is closed at time t=0andtheswitcht=0 and the switchS_{2}iskeptopen.Atsomelatertime(t0)is kept open. At some later time \left(\mathrm{t}_{0}\right), the switch S1\mathrm{S}_{1} is opened and S2\mathrm{S}_{2} is closed. the behaviour of the current I as a function of time ' t\mathrm{t} ' is given by:

Options

  1. AA solution of Fe _ 2 ( SO _ 4 ) _ 3 is electrolyzed for ' x ' min with a current of 1 . 5A to deposit 0 . 3482g of Fe .
  2. BA solution of Fe _ 2 ( SO _ 4 ) _ 3 is electrolyzed for ' x ' min with a current of 1 . 5A to deposit 0 . 3482g of Fe .
  3. CA solution of Fe _ 2 ( SO _ 4 ) _ 3 is electrolyzed for ' x ' min with a current of 1 . 5A to deposit 0 . 3482g of Fe .
  4. DA solution of Fe _ 2 ( SO _ 4 ) _ 3 is electrolyzed for ' x ' min with a current of 1 . 5A to deposit 0 . 3482g of Fe .

Correct answer

A. A solution of Fe _ 2 ( SO _ 4 ) _ 3 is electrolyzed for ' x ' min with a current of 1 . 5A to deposit 0 . 3482g of Fe .

Step-by-step solution

In the circuit shown, the switch S_ 1 is closed at time t=0 and the switch S_ 2 is kept open. At some later time (t_ 0
The current will grow for the time t=0\mathrm{t}=0 to t=t0\mathrm{t}=\mathrm{t}_{0} and after that decay of current takes place.
Growth and decay of current is of exponential nature
i=i0(1−e−t/τ)→ during growth \mathrm{i}=\mathrm{i}_0\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{\tau}}\right) \rightarrow \text{ during growth }
i=imax⁡e−t/τ→ during decay \mathrm{i}=\mathrm{i}_{\max } \mathrm{e}^{-\mathrm{t} / \mathrm{\tau}} \rightarrow \text{ during decay }
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