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

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 = L / R will be equal to <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/993e5ca4412eaf20245edde293ef9e13-.png" /> (A)  (e/ 1 - e ) (B) 1 (C) - 1 (D)  ( 1 - e / e)

Tardigrade Admin · Accepted Answer

Applying Kirchhoff's law in closed loop, -VR-VC=0 ⇒ VR / VC=-1 Note: The sense of voltage drop has not been defined. The answer could have been 1

$\frac{e}{1 - e}$

1

- 1

$\frac{1 - e}{e}$

Applying Kirchhoff's law in closed loop, $- V_{R} - V_{C} = 0$
$\Rightarrow V_{R} / V_{C} = - 1$
Note : The sense of voltage drop has not been defined. The answer could have been 1

1−ee​

1

- 1

e1−e​

Applying Kirchhoff's law in closed loop, −VR​−VC​=0 ⇒VR​/VC​=−1 Note : The sense of voltage drop has not been defined. The answer could have been 1

$\frac{e}{1 - e}$

$\frac{1 - e}{e}$

Applying Kirchhoff's law in closed loop, $- V_{R} - V_{C} = 0$
$\Rightarrow V_{R} / V_{C} = - 1$
Note : The sense of voltage drop has not been defined. The answer could have been 1