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 $t = 0$. The ratio of the voltage across the resistance $(V_R)$ and the inductor $(V_L)$ at $t = L/R$ will be equal to:

The circuit shown consists of a switch $S$,a battery $B$ of emf $E$,a resistance $R$,and an inductor $L$. What is the current in the circuit at the instant the switch $S$ is closed?

In the circuit shown below,the switch is closed at time $t=0$. Which of the graphs shown below best represents the voltage across the inductor,as seen on an oscilloscope?

Consider an electrical circuit containing a two-way switch $S$. Initially,$S$ is open and then $T_{1}$ is connected to $T_{2}$. As the current in $R = 6 \, \Omega$ attains a maximum value of steady-state level,$T_{1}$ is disconnected from $T_{2}$ and immediately connected to $T_{3}$. The potential drop across the $r = 3 \, \Omega$ resistor immediately after $T_{1}$ is connected to $T_{3}$ is $.... \, V$. (Round off to the nearest integer)

$A$ blackbox $(BB)$ which may contain a combination of electrical circuit elements (resistor, capacitor or inductor) is connected with other external circuit elements as shown in figure $(A)$. After the switch $S$ is closed at time $t=0$, the current $I$ as a function of time $t$ is shown in figure $(B)$. From this, we can infer that the blackbox contains:

An inductance coil has a time constant of $4 \, sec$. If it is cut into two equal parts and connected in parallel,then the new time constant of the circuit is.....$sec$.