Question

An RLcircuit consists of a $40.0\Omega$resistor and a 3.00 mHinductor. (a) Find its impedance Z at 60.0 Hzand 10.0 kHz. (b) Compare these values of Zwith those found in Example 23.12 in which there was also a capacitor.

Short answer

a) As a result, low and high-frequency impedance are respectively $188.5\Omega$and$193\Omega$

b) At high frequencies, the capacitor has a smaller influence, while having a larger influence at lower frequencies.

Step-by-step solution

Definition of circuit

A circuit is a closed path via which electricity can flow from one location to another. It could include transistors, resistors, and capacitors, among other electrical components.

Given data

Resistance in the RL circuit is $R=40.0\Omega$

The inductor in the RL circuit is$L=3.00mH\left(\frac{{10}^{-3}H}{1mH}\right)=3.00\times {10}^{-3}H$

Finding Impedance(a)

The formula for the determination of impedance of inductance can be expressed as,

$X_L=2\pi fL$………..(1)

Here $X_L$is inductive reactance, f is the frequency, and L is the inductance.

For each frequency, impedance can be expressed as,

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

………………(2)

At low frequency, f = 60 Hz, the value of $X_C=0\Omega$.

Substituting the given values in equation (1), we get

$$\begin{gathered} X_L=2\times 3.14\times \left(60Hz\right)\left(3.00\times {10}^{-3}H\right) \\ =1.13\Omega \end{gathered}$$

………………(3)

Then, using the given data and value from equation (3), the impedance can be calculated using equation (2), such that,

$$\begin{gathered} Z=\sqrt{\left({\left(40\Omega \right)}^2+{\left(1.13\Omega -0\Omega \right)}^2\right)} \\ =40.02\Omega \end{gathered}$$

At high-frequency $f=10.0kHz\left(\frac{{10}^{3}Hz}{1kHz}\right)=1.00\times {10}^{4}Hz$, substituting the given values in the above equation (1), we get

$$\begin{gathered} X_L^{\text{'}}=2\times 3.14\times \left(1\times {10}^{4}Hz\right)\left(3\times {10}^{-3}H\right) \\ =188.5\Omega \end{gathered}$$

Then, using the given data and value from equation (3), the impedance can be calculated using equation (2), such that,

$$\begin{gathered} Z^{\text{'}}=\sqrt{{\left({\left(40\Omega \right)}^2+\left(188.5\Omega \right)-0\Omega \right)}^2} \\ =192.7\Omega \end{gathered}$$

Therefore, the impedance at low and high frequencies are $188.5\Omega$and $193\Omega$.

Step 3: Compare these values of Z(b)

Compare these values of Z with those found in the example 23.12 , in which there was also a capacitor.

As we can see, at f =60Hz , with a capacitor the value of impedance is, $Z=531\Omega$, that is about 13 times as high as without the capacitor.

The capacitor makes a large difference at low frequencies.

At f = 10 kHz , with a capacitor, the value of impedance is, $Z=190\Omega$, that is similar to the value obtained without the capacitor.

Thus, the capacitor has a smaller effect at high frequencies.