Question
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 $$t = \frac{L}{R}$$ will be
equal to:
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 $$t = \frac{L}{R}$$ will be
equal to:
A.
$$\frac{e}{{1 - e}}$$
B.
1
C.
-1
D.
$$\frac{{1 - e}}{e}$$
Answer :
-1
Solution :
Applying Kirchhoff's law of voltage in closed loop
$$ - {V_R} - {V_C} = 0 \Rightarrow \frac{{{V_R}}}{{{V_C}}} = - 1$$
Applying Kirchhoff's law of voltage in closed loop
$$ - {V_R} - {V_C} = 0 \Rightarrow \frac{{{V_R}}}{{{V_C}}} = - 1$$
