In the circuit shown below, the key $$K$$ is closed at $$t=0.$$ The current through the battery is

An inductor of inductance $$L=400$$ $$mH$$ and resistors of resistance $${R_1} = 2\Omega $$ and $${R_2} = 2\Omega $$ are connected to a battery of $$emf$$ $$12$$ $$V$$ as shown in the figure. The internal resistance of the battery is negligible. The switch $$S$$ is closed at $$t=0.$$ The potential drop across $$L$$ as a function of time is :

In an $$a.c.$$ circuit the voltage applied is $$E = {E_0}\,\sin \,\omega t.$$ The resulting current in the circuit is $$I = {I_0}\sin \left( {\omega t - {\pi \over 2}} \right).$$ The power consumption in the circuit is given by

In a series resonant $$LCR$$ circuit, the voltage across $$R$$ is $$100$$ volts and $$R = 1\,k\Omega $$ with $$C = 2\mu F.$$ The resonant frequency $$\omega $$ is $$200$$ $$rad/s$$. At resonance the voltage across $$L$$ is