Question

An LC circuit consists of a\(3.00{\rm{ }}mH\)inductor and a\(5.00{\rm{ }}\mu F\)capacitor.

(a) Find its impedance at\(60.0{\rm{ }}Hz\)and\(10.0{\rm{ }}kHz\)

(b) Compare these values of\(Z\)with those found in Example\(23.12\)in which there was also a resistor.

Short answer

a) The impedance value at\(60{\rm{ }}Hz\)is:\(530.5\Omega \)and\(10kHz\)is:\(188.5\Omega \).

b) After comparing the values of \(Z\) we found that the impedance values are almost identical.

Step-by-step solution

Definition of circuit

An electrical circuit is a closed channel that allows electricity to go from one place to another. It might also have additional electrical components including resistors, capacitors, and transistors.

Given

Inductance,\(L = 3mH = 3 \times {10^{ - 3}}H\)

Capacitance,\(C = 5\mu F = 5 \times {10^{ - 6}}\;F\)

Frequency,

\(\begin{aligned} {f_1} &= 60\;Hz,{f_2}\\ &= 10kHz\\ &= 10 \times {10^3}\;Hz\end{aligned}\)

Calculation for Impedance

a)

Let us consider the given information,

Formula used:

The capacitive resistance is given as

\({X_C} = \frac{1}{{2\pi fC}}\)

The inductive resistance is given as

\({X_L} = 2\pi fL\)

Impedance of the LC circuit is given as

\(Z = \sqrt {{{\left( {{X_L} - {X_C}} \right)}^2}} \)

Here,

\(f\) is frequency

\(C\) is capacitance

\(L\) is inductance

Calculation:

The capacitive resistance is given as

\({X_C} = \frac{1}{{2\pi fC}}.......{\rm{ }}(1)\)

The inductive resistance is given as

\({X_L} = 2\pi fL\;\;\; \ldots \ldots ..{\rm{ }}(2)\)

Impedance of the\(LC\)circuit is given as

\(Z = \sqrt {{{\left( {{X_L} - {X_C}} \right)}^2}} \)

Plugging the values of Eq.\((1)\)and Eq.\((2)\)in the above expression

\(Z = \sqrt {{{\left( {(2\pi fL) - \left( {\frac{1}{{2\pi fC}}} \right)} \right)}^2}} \)

Plugging the values in the above equation

\(\begin{aligned} Z &= \sqrt {{{\left( {\left( {2\pi \times 60 \times 3 \times {{10}^{ - 3}}} \right) - \left( {\frac{1}{{2\pi \times 60 \times 5 \times {{10}^{ - 6}}}}} \right)} \right)}^2}} \\ &= 530.5\Omega \end{aligned}\)

The impedance of the\({\rm{LC}}\)circuit at\(10{\rm{kHz}}\)is calculated as

\(\begin{aligned} Z &= \sqrt {{{\left( {\left( {2\pi \times 10 \times {{10}^3} \times 3 \times {{10}^{ - 3}}} \right) - \left( {\frac{1}{{2\pi \times 10 \times {{10}^3} \times 5 \times {{10}^{ - 6}}}}} \right)} \right)}^2}} \\ &= 188.5\Omega \end{aligned}\)

Therefore, the value of impedance is \(188.5\Omega \) and \(530.5\Omega \).

Comparing the impedance values to the value obtained in the RC circuit

b)

Evaluating further,

The impedance values are really similar, because the capacitor dominates at low frequencies and the inductor dominates at higher frequencies, the values of impedance calculated in Example\(23.12\),\(Z = 531\Omega \) at\(60{\rm{ }}Hz\) and \(Z = 190\Omega \)at\(10kHz\), are very close to the values obtained in part (a).

This is because the capacitor dominates at low frequencies and the inductor dominates at higher frequencies, there is very little effect of the resistor in the circuit. The impedance values are really similar.

Because the capacitor dominates at low frequencies and the inductor dominates at higher frequencies, the value of impedance calculated in example\(23.12\).\(Z = 531\Omega \) at\(60{\rm{ }}Hz\)and\(Z = 190\Omega \)at\(10kHz\), which is very close to the value of impedance obtained in part (a). This is because the capacitor dominates at low frequencies and the inductor dominates at higher frequencies, there is very little effect of the resistor in the circuit.

Therefore, the impedance values are really similar.