### AC Impedance

• Calculate the impedance, phase angle, power, power factor, voltage, and/or current in a RLC series circuit consisting of any combination of resistors, capacitors, and inductors.

• Express total impedance in rectangular and polar forms,

• Explain the relationship between resistance, reactance, and impedance as it relates to the rectangular form of total impedance,

• Sketch an impedance triangle.

Voltage in an AC circuit is given by: $V = V_{max}\ \sin \omega t$

While calculating Impedance (equivalent of resistance in AC circuits)

• Resistance is plotted on x-axis as it is, going towards right direction.

• Inductance is calculated using the following formula and plotted vertically upwards. $X_{L} = 2 \pi f L$

• Capacitance is calculated using the following formula and plotted vertically downwards. $X_{C} = \frac{1}{2 \pi f C}$

So, the combined effect of inductance and capacitance Reactance is plotted on y-axis by taking the difference of $\displaystyle X_{L}$ and $\displaystyle X_{C}$. $X = X_L - X_C$ Impedance is calculated using Pythagoras theorem $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Power angle $\phi$ is the angle between Z and R. Power factor = cos$\phi$.

• For purely resistive circuit, $\phi$ = 0 and power factor = 1.

• For purely inductive circuit, $\phi$ = 90 and power factor = 0.

• For purely capacitive circuit, $\phi$ = -90 and power factor = 0.

### Solved Example:

#### 20-1-01

A 10 $\mu$F capacitor is connected across a 10 Volt battery, the steady state current is:

Solution:
For a capacitor, $i_c(t)=C * \dfrac{dv_c(t)}{dt}$ When switch is closed there is voltage difference between the positive and negative terminal. Hence, there will be flow of current until capacitor is charged. Since battery is a constant voltage source, the capacitor will be charged to 10 V and after that there won't be any flow of DC current through it and it will act as open circuit for DC current.

### Solved Example:

#### 20-1-02

When the frequency of the voltage applied to a series RC circuit is increased, the phase angle:

Solution:
In AC circuits, capacitive reactance is calculated using the following formula and plotted vertically downwards. $X_C = \dfrac{1}{2 \pi f C}$ Since the capacitive reactance is inversely proportional to the frequency, as frequency increases, capacitive reactance decreases. Power triangle becomes smaller as the vertical side of the power triangle decreases. Hence, phase angle decreases.

### Solved Example:

#### 20-1-03

A 470 $\Omega$ resistor and a coil with 125 $\Omega$ inductive reactance are in parallel. Both components are across a 15 V ac voltage source. Current through the inductor is:

Solution:
Since there is no capacitor, we can directly calculate the current. The inductance will be connected directly to 15V because it is a parallel circuit. Current = $\dfrac{15}{125}$ = 0.12 A = 120 mA

### Solved Example:

#### 20-1-04

To increase the current in a series RL circuit, the frequency:

Solution:
The value of resistance is independent of frequency. However, Inductance $X_{L} = 2 \pi f L$ is directly proportional to the frequency, due to which Impedance increases with increase in frequency.
As frequency increases, Impedance increases, and current decreases. To avoid this, frequency must be decreased to get higher values of current.

### Solved Example:

#### 20-1-05

A 12 k$\Omega$ resistor is in series with a 90 mH coil across an 8 kHz AC source. Current flow in the circuit, expressed in polar form, is I = 0.30 $\angle$ 0 mA. The source voltage, expressed in polar form, is:

Solution:
$f=8kHz$, $i=0.30\times 10^{-3} \angle 0,$
$X_{L} =2\pi fL =2\pi \left( 8000\right) \left( 90\times 10^{-3}\right) = 4.524\ k\Omega$ $Z =\sqrt {R^{2}+X_{L}^{2}} =\sqrt {12^{2}+4.524^{2}} =12.824\ k\Omega$ $\phi =\tan ^{-1}\left( \dfrac {X_{L}}{R}\right)\\ =\tan ^{-1}\left( \dfrac {4.524}{12}\right) =20.6^\circ$ \begin{align*} Z &=\dfrac {V}{i} \\ 12.824 \angle 20.6^\circ &=\dfrac {V}{0.30\times 10^{-3} \angle 0} \\ V &=3.85 \angle 20.6^\circ \end{align*}

### Solved Example:

#### 20-1-06

A 1.5 k$\Omega$ resistor and a coil with a 2.2 k$\Omega$ inductive reactance are in series across an 18 V AC source. The power factor is:

Solution:
$\tan \phi = \dfrac{2.2}{1.5}, \quad \phi = \tan^{-1} \left(\frac{2.2}{1.5} \right) = 55.7^\circ$ $\mathrm{Power\ Factor} = \cos \phi = \cos (55.7^\circ) = 0.564$

### Solved Example:

#### 20-1-07

In a series circuit, R = 300 $\Omega$, L = 0.9 H, C = 2.0 $\mu$ F and $\omega$ = 1000 rad/sec. The impedance of the circuit is: (MHT-CET 2019)

Solution:
$X_L = \omega L = (1000)\times (0.9)=900\ \Omega$ $X_C = \dfrac{1}{\omega C} = \dfrac{1}{1000 \times 2\times 10^{-6}}= 500\ \Omega$ $Z = \sqrt{R^2 + (X_L - X_C)^2} \\ = \sqrt{300^2 + 400^2} = 500\ \Omega$

### Solved Example:

#### 20-1-08

When 100 volts DC is supplied across a solenoid (Resistance and inductor in series), a current of 1.0 A flows in it. When 100 volts AC is applied across the same coil, the current drops to 0.5 A. If the frequency of AC source is 50 Hz, then the impedance and inductance of the solenoid are:

Solution:
For DC, Inductor part will not be effective. $R = \dfrac{V}{i} = 100\ \Omega$ For AC, $Z = \dfrac{V}{i} = \dfrac{100}{0.5} = 200\ \Omega$ \begin{align*} Z &= \sqrt{R^2 + (X_L)^2}\\ 200^2 &= 100^2 + (X_L)^2\\ X_L &= 173.2\ \Omega \end{align*} Given f = 50 Hz and Using, $X_L = 2 \pi f L$ $L = 0.55\ H$

### Solved Example:

#### 20-1-09

The potential difference V and the current i flowing through an instrument in an AC circuit of frequency f are given by V=5 $\cos \omega$t volts and I = 2 $\sin \omega$t amperes (where $\omega$ = 2$\pi$f). The power dissipated in the instrument is:

Solution:
$V=5 \cos \omega t=5 \sin( \omega t + \dfrac{\pi}{2})$ and $i=2 \sin \omega t$ $\mathrm{Power} = V_{RMS} \times i_{RMS} \times \cos \phi = 0$ (Since $\phi = \dfrac{\pi}{2}$, therefore $\cos \phi =\cos \dfrac{\pi}{2}$=0)

### Solved Example:

#### 20-1-10

A resistance R and a capacitor C connected in series across a 250V supply draw 5A at 50Hz. When frequency is increased to 60Hz, it draws 5.8A. Find the value of C. (PHCET 2018)

Solution:
Case I: When f = 50 Hz
$Z=\dfrac{V}{I} = \dfrac {250}{5}=50\ \Omega$ $X_{C}=\dfrac {1}{2\pi fC}=\dfrac {1}{2\pi \left( 50\right) C}=\left( \dfrac {3.18\times 10^{-3}}{C}\right) \Omega$ \begin{align*} Z^{2}&=R^{2}+{X_C}^{2}\\ 2500&=R^{2}+\left( \dfrac {3.18\times 10^{-3}}{C}\right)^{2} \end{align*} Case II: When f = 60 Hz $X_{C}=\dfrac {1}{120\pi C}= \dfrac{2.65\times 10^{-3}}{C}\ \Omega$ $Z=\dfrac {250}{5.8}=43.10\ \Omega$ $1857.9=R^{2}+\left( \dfrac {2.65\times 10^{-3}}{C}\right) ^{2}$ Subtracting I-II \begin{align*} 642.09 &=\dfrac {3.09\times 10^{-6}}{C^{2}}\\ C^{2} &=4.812\times 10^{-9}\\ C &=6.94\times 10^{-5}\ F =69.4\ \mu F \end{align*}

### Solved Example:

#### 20-1-11

If the load impedance is 20 - j20, the power factor is: (RSMSSB JE Electrical Engineering May 2022)

Solution:
Impedance $Z = \sqrt{R^2 + X^2}$ $Z = \sqrt{20^2 + (-20)^2} = 20\sqrt{2} \Omega$ Power factor is: $\cos \phi = \dfrac{R}{Z} = \dfrac{20}{20\sqrt{2}} = 0.707$

The phase voltage and current across a load element are 100.0 $\angle$ 45$^\circ$ V and 5.0 $\angle$15$^\circ$ A. respectively. Determine the impedance and admittance of the load. (SSC JE EE March 2021 Evening)
$Z = \dfrac{V}{I} = \dfrac{100 \angle 45}{5 \angle 15} = 20 \angle 30^\circ$ $Y ==\dfrac{1}{Z} = \dfrac{1}{20\angle 30} 0.05\angle -30^\circ$