Question

A resistor of resistance R, capacitor C and inductor of inductance L are connected parallel to AC power source of voltage ${\varepsilon _o}\sin \omega t$. The maximum current through the resistance is half of the maximum current through the power source. Then the value of R is:

Answer 1

Hint: Capacitor is defined as the device that stores electrical energy in any electric field. Represented by C and written in terms of Faraday (F). The insulator is defined as the device that stores energy in a magnetic field if current is flown through the inductor and unit of insulator is Henry (H).

Formula used:
The formula of the resistance is given by,
$ \Rightarrow {\varepsilon _o} = {i_{Ro}} \times R$
Where potential difference is ${\varepsilon _o}$, the current is ${i_{Ro}}$ and resistance is R.
The formula for the inductor is given by,
$ \Rightarrow {\varepsilon _o} = {i_{Lo}} \times \omega L$
Where potential difference is ${\varepsilon _o}$, the current is ${i_{Lo}}$ and resistance is $\omega L$.
The formula of the capacitor is given by,
$ \Rightarrow {\varepsilon _o} = {i_{Co}} \times \dfrac{1}{{\omega C}}$
Where potential difference is ${\varepsilon _o}$, the current is ${i_{Co}}$ and the resistance is $\dfrac{1}{{\omega C}}$.

Complete step by step solution:
A resistor of resistance R, capacitor C and inductor of inductance L are connected parallel to AC power source of voltage ${\varepsilon _o}\sin \omega t$ and the maximum current through the resistance is half of the maximum current through the power source then we need to find the value of R.
As it is given that the maximum current through the resistance is half of the maximum current through the power source, therefore.
$ \Rightarrow {i_{Ro}} = \dfrac{1}{2} \times \sqrt {{{\left( {{i_{Ro}}} \right)}^2} + {{\left( {{i_{Co}} - {i_{Lo}}} \right)}^2}} $.
Replacing the values of current of the resistance, capacitor and inductor we get.
$ \Rightarrow \dfrac{{{\varepsilon _o}}}{R} = \dfrac{1}{2} \times \sqrt {{{\left( {\dfrac{{{\varepsilon _o}}}{R}} \right)}^2} + {{\left( {{\varepsilon _o}\omega C - \dfrac{{{\varepsilon _o}}}{{\omega L}}} \right)}^2}} $
The ${\varepsilon _o}$ is the voltage from the power source and the angular frequency of the voltage source is $\omega $.
$ \Rightarrow \dfrac{{{\varepsilon _o}}}{R} = \dfrac{1}{2} \times \sqrt {\dfrac{{\varepsilon _o^2}}{{{R^2}}} + {{\left( {{\varepsilon _o}\omega C} \right)}^2} + {{\left( {\dfrac{{{\varepsilon _o}}}{{\omega L}}} \right)}^2} - \dfrac{{2\varepsilon _o^2C}}{L}} $
$ \Rightarrow R = \dfrac{{\sqrt 3 }}{{\left( {\omega C - \dfrac{1}{{\omega L}}} \right)}}$.

The resistance is equal to $R = \dfrac{{\sqrt 3 }}{{\left( {\omega C - \dfrac{1}{{\omega L}}} \right)}}$.

Note: If it is advised for the students to remember and understand the formula of the relation between the current, voltage and resistance for inductor, resistor and capacitor as it is very helpful for solving problems like these.