Question

An ac source of angular frequency \[\omega \] is fed across a resistor \[R\] and a capacitor \[C\] in series. The current registered is \[I\]. If now the frequency of the source is changed to \[\omega /3\] (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of the reactance to resistance at the original frequency \[\omega \] is given as \[\sqrt {\dfrac{x}{5}} \]. Find \[x\].

Answer 1

Hint: Use the formula for the root mean square current. This formula gives the relation between the root mean square voltage, resistance and capacitive reactance. Also use the formula for capacitive reactance in terms of the angular frequency and capacitance. Write these equations for the given two conditions of current and angular frequency and determine the ratio of the reactance to resistance.

Formulae used:
The root mean square current \[{I_{rms}}\] in the RC circuit is given by
\[{I_{rms}} = \dfrac{{{V_{rms}}}}{{\sqrt {{R^2} + X_C^2} }}\] …… (1)
Here, \[{V_{rms}}\] is the root mean square voltage, \[R\] is the resistance and \[{X_C}\] is the capacitive reactance.
The capacitive reactance \[{X_C}\] is given by
\[{X_C} = \dfrac{1}{{\omega C}}\] …… (2)
Here, \[\omega \] is the angular frequency and \[C\] is the capacitance.

Complete step by step answer:
We have given that the initial angular frequency of the source is \[\omega \] and the current is \[I\].
Substitute \[I\] for \[{I_{rms}}\] and \[\dfrac{1}{{\omega C}}\] for \[{X_C}\] in equation (1).
\[I = \dfrac{{{V_{rms}}}}{{\sqrt {{R^2} + {{\left( {\dfrac{1}{{\omega C}}} \right)}^2}} }}\]
\[ \Rightarrow I = \dfrac{{{V_{rms}}}}{{\sqrt {{R^2} + \dfrac{1}{{{\omega ^2}{C^2}}}} }}\]
\[ \Rightarrow I = \dfrac{{{V_{rms}}\omega C}}{{\sqrt {{R^2}{\omega ^2}{C^2} + 1} }}\] …… (3)
When the angular frequency of the source is changed to \[\dfrac{\omega }{3}\] then the current becomes \[\dfrac{I}{2}\] but the potential is the same. Hence, the above equation becomes
\[ \Rightarrow \dfrac{I}{2} = \dfrac{{{V_{rms}}\dfrac{\omega }{3}C}}{{\sqrt {{R^2}{{\left( {\dfrac{\omega }{3}} \right)}^2}{C^2} + 1} }}\]
\[ \Rightarrow \dfrac{I}{2} = \dfrac{{{V_{rms}}\omega C}}{{\sqrt {{R^2}{\omega ^2}{C^2} + 9} }}\] …… (4)

Divide equation (3) by equation (4).
\[ \Rightarrow \dfrac{I}{{\dfrac{I}{2}}} = \dfrac{{\dfrac{{{V_{rms}}\omega C}}{{\sqrt {{R^2}{\omega ^2}{C^2} + 1} }}}}{{\dfrac{{{V_{rms}}\omega C}}{{\sqrt {{R^2}{\omega ^2}{C^2} + 9} }}}}\]
\[ \Rightarrow 2 = \dfrac{{\sqrt {{R^2}{\omega ^2}{C^2} + 9} }}{{\sqrt {{R^2}{\omega ^2}{C^2} + 1} }}\]

Take square on both sides of the above equation.
\[ \Rightarrow 4 = \dfrac{{{R^2}{\omega ^2}{C^2} + 9}}{{{R^2}{\omega ^2}{C^2} + 1}}\]
\[ \Rightarrow 4{R^2}{\omega ^2}{C^2} + 4 = {R^2}{\omega ^2}{C^2} + 9\]
\[ \Rightarrow 4{R^2}{\omega ^2}{C^2} - {R^2}{\omega ^2}{C^2} = 9 - 4\]
\[ \Rightarrow 3{R^2}{\omega ^2}{C^2} = 5\]
\[ \Rightarrow {R^2}{\omega ^2}{C^2} = \dfrac{5}{3}\]
\[ \Rightarrow {R^2} = \dfrac{5}{3}\dfrac{1}{{{\omega ^2}{C^2}}}\]

Take square root on both sides of the above equation.
\[ \Rightarrow R = \sqrt {\dfrac{5}{3}} \dfrac{1}{{\omega C}}\]
Substitute \[{X_C}\] for \[\dfrac{1}{{\omega C}}\] in the above equation.
\[ \Rightarrow R = \sqrt {\dfrac{5}{3}} {X_C}\]
\[ \therefore \dfrac{{{X_C}}}{R} = \sqrt {\dfrac{3}{5}} \]

Hence, the ratio of reactance to resistance at the original frequency is \[\sqrt {\dfrac{3}{5}} \]. Therefore, the value of x is 3.

Note: One can also solve the same question by using the same formulae but with different calculations. One can equate the root mean square potential or voltage for the two different conditions of the current and angular frequency and determine the ratio of the reactance to resistance at original frequency.