An inductor $$(L=0.03$$ $$H)$$ and a resistor $$\left( {R = 0.15\,k\Omega } \right)$$ are connected in series to a battery of $$15V$$ $$EMF$$ in a circuit shown below. The key $${K_1}$$ has been kept closed for a long time. Then at $$t=0$$, $${K_1}$$ is opened and key $${K_2}$$ is closed simultaneously. At $$t=1$$ $$ms,$$ the current in the circuit will be : $$\left( {{e^5} \cong 150} \right)$$

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 :

In an $$LCR$$ circuit as shown below both switches are open initially. Now switch $${S_1}$$ is closed, $${S_2}$$ kept open. ($$q$$ is charge on the capacitor and $$\tau $$ $$=RC$$ is Capacitance time constant). Which of the following statement is correct ?

A resistor $$'R'$$ and $$2\mu F$$ capacitor in series is connected through a switch to $$200$$ $$V$$ direct supply. Across the capacitor is a neon bulb that lights up at $$120$$ $$V.$$ Calculate the value of $$R$$ to make the bulb light up $$5$$ $$s$$ after the switch has been closed. $$\left( {{{\log }_{10}}2.5 = 0.4} \right)$$