Question

An inductance of 2.0 H, a capacitance of 18μF and a resistance of 10 kΩ are connected to an AC source of 20 V with adjustable frequency. (a) What frequency should be chosen to maximise the current in the circuit? (b) What is the value of this maximum current?

Answer 1

a) The frequency that should be chosen to maximise the current in the circuit is 27 Hz b)The value of the maximum current is 2 mA

Explanation:

Inductor of Inductance, L = 2.0 H

Capacitor of Capacitance , C = 18 μF    

Resistor of Resistance , R = 10 kΩ

Source voltage AC, E = 20 V

(a) In an LCR circuit, the current is maximum when the reaction is minimal, which happens at the resonance, i.e. when the inductive reaction is equal to the capacitive reaction, i.e.

$\begin{aligned}X_{L} &=X_{C} \\\omega L &=\frac{1}{\omega C} \\\omega^{2}=\frac{1}{L C} &=\frac{1}{\left(2 \times 18 \times 10^{-6}\right)} \\\omega^{2} &=\frac{10^{6}}{36} \\2 \pi f &=\frac{10^{3}}{6} \\f=& \frac{1000}{(6 \times 2 \pi)}\end{aligned}$

          $\begin{aligned}=& 26.539 \mathrm{Hz} \\=&27 \mathrm{Hz}\end{aligned}$

(b) Reactance is small at resonance.

Small Reactance, Z = R

It gives maximum current (I) by,

$\begin{aligned}I &=\frac{E}{R} \\I=& \frac{20}{\left(10 \times 10^{3}\right)} \\I=& \frac{2 A}{10^{3}} \\=&2 \mathrm{mA}\end{aligned}$