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) :

A $$10 \mu \mathrm{F}$$ capacitor is connected to a $$210 \mathrm{~V}, 50 \mathrm{~Hz}$$ source as shown in figure. The peak current in the circuit is nearly $$(\pi=3.14)$$ :

An ac source is connected in the given circuit. The value of $\phi$ will be :