JEE MAIN - Physics (2014 (Offline) - No. 7)
In the circuit shown here, the point $$'C'$$ is kept connected to point $$'A'$$ till the current flowing through the circuit becomes constant. Afterward, suddenly, point $$'C'$$ is disconnected from point $$'A'$$ and connected to point $$'B'$$ at time $$t=0.$$ Ratio of the voltage across resistance and the inductor at $$t=L/R$$ will be equal to :
_en_7_1.png)
$${e \over {1 - e}}$$
$$1$$
$$-1$$
$${{1 - e} \over e}$$
Explanation
Applying kirchhoffs law of voltage in closed loop
$$ - {V_R} - {V_C} = 0 \Rightarrow {{{V_R}} \over {{V_C}}} = - 1$$
_en_7_2.png)
$$ - {V_R} - {V_C} = 0 \Rightarrow {{{V_R}} \over {{V_C}}} = - 1$$
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