$A$ coil of inductance $0.20 \, H$ is connected in series with a switch and a cell of $emf$ $1.6 \, V$. The total resistance of the circuit is $4.0 \, \Omega$. What is the initial rate of growth of the current when the switch is closed? (in $A \, s^{-1}$)

The ratio of the time constant in charging and discharging in the circuit shown in the figure is

$A$ $10 \text{ cm}$ long perfectly conducting wire $PQ$ is moving with a velocity $1 \text{ cm/s}$ on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor $L = 1 \text{ mH}$ and a resistance $R = 1 \ \Omega$ as shown in the figure. The horizontal rails,$L$,and $R$ lie in the same plane with a uniform magnetic field $B = 1 \text{ T}$ perpendicular to the plane. If the key $S$ is closed at a certain instant,the current in the circuit after $1 \text{ millisecond}$ is $x \times 10^{-3} \text{ A}$,where the value of $x$ is. . . . . . [Assume the velocity of wire $PQ$ remains constant $(1 \text{ cm/s})$ after key $S$ is closed. Given: $e^{-1} = 0.37$,where $e$ is the base of the natural logarithm]

An emf of $15 V$ is applied to a circuit containing $5 H$ inductance and $10 \Omega$ resistance. The ratio of currents at time $t=\infty$ and $t=1 s$ after closing the switch is:

Find the current in the sliding rod $AB$ (resistance $= R$) for the arrangement shown in the figure. $\vec{B}$ is constant and is directed out of the paper. The parallel wires have no resistance. $v$ is constant. The switch $S$ is closed at time $t = 0$.

In the circuit shown, when the current $i$ is $3 \text{ A}$ and increasing at the rate of $1 \text{ A/s}$, the potential difference between $A$ and $B$ is $12 \text{ V}$. When the same current $3 \text{ A}$ is decreasing at the rate of $1 \text{ A/s}$, the potential difference between $A$ and $B$ is $6 \text{ V}$. The value of $R$ in the circuit is: (in $\text{ } \Omega$)