In a series $$LCR$$ circuit $$R = 200\Omega $$ and the voltage and the frequency of the main supply is $$220V$$ and $$50$$ $$Hz$$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by $${30^ \circ }.$$ On taking out the inductor from the circuit the current leads the voltage by $${30^ \circ }.$$ The power dissipated in the $$LCR$$ circuit is

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