Question

A series combination of resistor \[\left( R \right)\] and the capacitor \[\left( C \right)\] is connected to an AC source of angular frequency \['\omega '\]. Keeping the voltage the same, if the frequency is changed to \[\dfrac{\omega }{3}\] , the current becomes half of the original current. Then the ratio of the capacitive reactance and resistance at the former frequency is:
A. \[\sqrt {0.6} \]
B. \[\sqrt 3 \]
C. \[\sqrt 2 \]
D. \[\sqrt 6 \]

Answer 1

Hint: Here, we have been given a combination series of resistor and a capacitor and are asked to calculate the ratio of the capacitive reactance and the resistance at current frequency. Thus we will use the R.M.S. value of supply voltage, impedance of RC circuit.

Complete step by step answer:
Let R.M.S. value of supply voltage be \[v\] . The impedance of RC circuit is given by:
\[Z = \sqrt {{R^2} + X_C^2} \] ; \[Z\] , \[R\] and \[{X_C}\] represents impedance, resistance and capacitive reactance respectively. It is also represented as:
\[Z = \sqrt {{R^2} + X_C^2} = \sqrt {{R^2} + \dfrac{1}{{{\omega ^2}{C^2}}}} \]
So the R.M.S. current in the circuit is given by the original current in the circuit as
\[{I_1} = \dfrac{v}{Z}\]
\[ \Rightarrow {I_1} = \dfrac{v}{{\sqrt {{R^2} + \dfrac{1}{{{\omega ^2}{C^2}}}} }}\]
As we know that when keeping the voltage same, if the frequency is changed to \[\dfrac{\omega }{3}\] , the current becomes half of the original current i.e. \[{I_2} = \dfrac{1}{2}{I_1}\]
Therefore,
\[\dfrac{{{I_2}}}{{{I_1}}} = \dfrac{1}{2}\]...................…. (Given)
 \[\Rightarrow \dfrac{{{I_2}}}{{{I_1}}} = \dfrac{{\sqrt {{R^2} + \dfrac{1}{{{\omega ^2}{C^2}}}} }}{{\sqrt {{R^2} + \dfrac{1}{{{{\left( {\dfrac{\omega }{3}} \right)}^2}{C^2}}}} }}\] ; \[{\omega _2} = \dfrac{\omega }{3}\] given in the question.
On squaring both the sides of the above equation gives:
\[{\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)^2} = \dfrac{{{R^2} + \dfrac{1}{{{\omega ^2}{C^2}}}}}{{{R^2} + \dfrac{1}{{{{\left( {\dfrac{\omega }{3}} \right)}^2}{C^2}}}}}\]
On substituting all the values and simplifying, we get
\[\dfrac{1}{4} = \dfrac{{{R^2}\left( {1 + {\omega ^2}{R^2}{C^2}} \right)}}{{{R^2}\left( {9 + {\omega ^2}{R^2}{C^2}} \right)}}\]
\[\Rightarrow \dfrac{1}{4} = \dfrac{{\left( {1 + {\omega ^2}{R^2}{C^2}} \right)}}{{\left( {9 + {\omega ^2}{R^2}{C^2}} \right)}}\]
\[ \Rightarrow \omega RC = \sqrt {\dfrac{5}{3}} \]
Ratio of the capacitive reactance and resistance in former frequency is:
\[\dfrac{{{X_C}}}{R} = \dfrac{1}{{\omega RC}} \]
$\Rightarrow \dfrac{{{X_C}}}{R} = \sqrt {\dfrac{3}{5}}$
\[\therefore \dfrac{{{X_C}}}{R} = \sqrt {0.6} \]
Thus the ratio is obtained as \[\sqrt {0.6} \].

Hence, the correct answer is option A.

Note: We have first used the formula for impedance in the RC circuit and then calculated the original current from that we have been able to calculate the final value of \[\omega RC\] which is the ratio between resistance and the capacitive reactance the inverse will be the ratio we required here. Be careful while replacing the values.