In the circuit shown below, the inductance $$L$$ is connected to an ac source. The current flowing in the circuit is $$I=I_0 \sin \omega t$$. The voltage drop $$\left(V_L\right)$$ across $$L$$ is

A step up transformer is connected to an ac mains supply of $$220 \mathrm{~V}$$ to operate at $$11000 \mathrm{~V}, 88$$ watt. The current in the secondary circuit, ignoring the power loss in the transformer, is

The amplitude of the charge oscillating in a circuit decreases exponentially as $$Q=Q_0 e^{-R t/2 L}$$, where $$Q_0$$ is the charge at $$t=0 \mathrm{~s}$$. The time at which charge amplitude decreases to $$0.50 Q_0$$ is nearly:

[Given that $$R=1.5 \Omega, L=12 \mathrm{~mH}, \ln (2)=0.693$$]

In an ideal transformer, the turns ratio is $$\frac{N_P}{N_S}=\frac{1}{2}$$. The ratio $$V_S: V_P$$ is equal to (the symbols carry their usual meaning) :