JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 18)
A 20 Henry inductor coil is connected to a
10 ohm resistance in series as shown in figure.
The time at which rate of dissipation of energy
(joule's heat) across resistance is equal to the
rate at which magnetic energy is stored in the
inductor is :
_8th_April_Morning_Slot_en_18_1.png)
$${2 \over {\ln 2}}$$
$${\ln 2}$$
$$2{\ln 2}$$
$${1 \over 2}{\ln 2}$$
Explanation
PR = i2 × R
PB = V × i
$$ \therefore $$ PL = Vi – i2R
$$ \Rightarrow $$ Vi – i2R = i2R
$$ \Rightarrow $$ $$i = {V \over {2R}}\,\,and\,\,i = {V \over R}\left( {1 - {e^{ - t/\tau }}} \right)$$
$$ \therefore $$ $${V \over {2R}} = {V \over R}\left( {1 - {e^{ - t/\tau }}} \right)$$
$$ \Rightarrow $$ t = $$\tau $$ln(2) = $${{20} \over {10}}ln(2) = 2ln(2)$$
PB = V × i
$$ \therefore $$ PL = Vi – i2R
$$ \Rightarrow $$ Vi – i2R = i2R
$$ \Rightarrow $$ $$i = {V \over {2R}}\,\,and\,\,i = {V \over R}\left( {1 - {e^{ - t/\tau }}} \right)$$
$$ \therefore $$ $${V \over {2R}} = {V \over R}\left( {1 - {e^{ - t/\tau }}} \right)$$
$$ \Rightarrow $$ t = $$\tau $$ln(2) = $${{20} \over {10}}ln(2) = 2ln(2)$$
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