KVPY Sample Paper KVPY Stream-SX Model Paper-11

An ac source of angular frequency \[\omega \] is fed 3 across a resistor r and a capacitor C in series. The current registered is \[I\]. If now the frequency Of source is changed to \[\omega /3\] (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency \[\omega \]

A) \[\sqrt{\frac{3}{5}}\]

B) \[\sqrt{\frac{2}{5}}\]

C) \[\sqrt{\frac{1}{5}}\]

D) \[\sqrt{\frac{4}{5}}\]

Correct Answer: A

Solution :

\[i=\frac{V}{\sqrt{{{R}^{2}}+{{\left( \frac{1}{\omega C} \right)}^{2}}}}\]
Or \[I=\frac{V}{\sqrt{{{R}^{2}}+\frac{1}{{{\omega }^{2}}{{C}^{2}}}}}\]	?.(i)
And =\[\frac{I}{2}=\frac{V}{\sqrt{{{R}^{2}}+\frac{{{\omega }^{2}}{{C}^{2}}}{9}}}\]	... (ii)
On simplifying above equation, we get \[\frac{{{X}_{L}}}{R}=\frac{\omega L}{R}=\sqrt{\frac{3}{5}.}\]