JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 17)
The current (i) at time t = 0 and t = $$\infty $$ respectively for the given circuit is :
_25th_February_Morning_Shift_en_17_1.png)
$${{18E} \over {55}},{{5E} \over {18}}$$
$${{5E} \over {18}},{{18E} \over {55}}$$
$${{5E} \over {18}},{{10E} \over {33}}$$
$${{10E} \over {33}},{{5E} \over {18}}$$
Explanation
_25th_February_Morning_Shift_en_17_2.png)
at t = 0, inductor is removed, so circuit will look like this
at t = 0
_25th_February_Morning_Shift_en_17_3.png)
$${R_{eq}} = {{6 \times 9} \over {6 + 9}} = {{54} \over {15}}$$
$$I(t = 0) = {{E \times 15} \over {54}} = {{5E} \over {18}}$$
at t = $$\infty$$, inductor is replaced by plane wire, so circuit will look like this
at t = $$\infty$$,
_25th_February_Morning_Shift_en_17_4.png)
$$I(t = \infty ) = {E \over {{5 \over 2} + {4 \over 5}}} = {{10E} \over {33}}$$
Now,
_25th_February_Morning_Shift_en_17_5.png)
$${R_{eq}} = {{1 \times 4} \over {1 + 4}} + {{5 \times 5} \over {5 + 5}}$$
$$ = {4 \over 5} + {5 \over 2} = {{8 + 25} \over {10}} = {{33} \over {10}}$$
$$I = {E \over {{R_{eq}}}} = {{10E} \over {33}}$$
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