Learning Objectives:

• 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:

A. 10 Amp

B. $10^6$ Amp

C. 1 Amp

D. Zero

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.

Correct Answer: D

Solved Example: 20-1-02

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

A. Decreases

B. Increases

C. Becomes erratic

D. Remains the same

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.

Correct Answer: A

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:

A. 32 mA

B. 120 mA

C. 12 mA

D. 152 mA

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

Correct Answer: B

Solved Example: 20-1-04

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

A. Should be constant

B. Cannot be determined without values

C. Should be increased

D. Should be decreased

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.

Correct Answer: D

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:

A. 3.84 $\angle$ 20.6 V

B. 0.32 $\angle$ 0.6 V

C. 12.8 $\angle$ 20.6 V

D. 13.84 $\angle$ 69.4 V

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

Correct Answer: A

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:

A. 0.922

B. 0.564

C. 0.76

D. 0.557

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

Correct Answer: B

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)

A. 900 $\Omega$

B. 500 $\Omega$

C. 400 $\Omega$

D. 1300 $\Omega$

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

Correct Answer: B

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:

A. 200$\Omega$, 0.55H

B. 100$\Omega$, 0.86H

C. 200$\Omega$, 1.0H

D. 100$\Omega$, 0.93H

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, \[X_L = 2 \pi f L\] \[L = 0.55\ H\]

Correct Answer: A

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:

A. Zero

B. 10 W

C. 5 W

D. 2.5 W

\[V=5 \cos \omega t=5 \sin( \omega t + \dfrac{\pi}{2})\] 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)

Correct Answer: A

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)

A. 69.4 $\mu$F

B. 122.4 $\mu$F

C. 220.7 $\mu$F

D. 371.9 $\mu$F

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*}

Correct Answer: A

Solved Example: 20-1-11

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

A. $45^\circ$

B. 0

C. 1

D. 0.707

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

Correct Answer: D

Solved Example: 20-1-12

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)

A. Z = 20.0 $\angle$-30$^\circ$ Ω and Y = 0.05$\angle$30$^\circ$ S

B. Z = 20.0 $\angle$30$^\circ$ Ω and Y = 0.05$\angle$-30$^\circ$ S

C. Z = 0.05 $\angle$30$^\circ$ Ω and Y = 20$\angle$-30$^\circ$ S

D. Z = 0.05$\angle$-30$^\circ$ Ω and Y = 20 $\angle$30$^\circ$ S

\[Z = \dfrac{V}{I} = \dfrac{100 \angle 45}{5 \angle 15} = 20 \angle 30^\circ\] \[Y ==\dfrac{1}{Z} = \dfrac{1}{20\angle 30} 0.05\angle -30^\circ\]

Correct Answer: B

Learning Objectives:

• Calculate the mean value of a function

Average Value: The arithmetic average of all the values of an alternating quantity over one cycle is called its average value.

\[X_{avg} =\dfrac{1}{T}\int_{0}^{T}x(t)dt\]

For Symmetrical waveforms, the average value calculated over one cycle becomes equal to zero because the positive area cancels the negative area. Hence for symmetrical waveforms, the average value is calculated for half cycle.

Solved Example: 20-2-01

For the same peak value, which of the following wave has the least mean value ?

A. Half wave rectified sine wave

B. Triangular wave

C. Sine wave

D. Square wave

Half wave rectified sine wave has zero value for half cycle corresponding to the negative half cycle of input waveform. Hence, it's mean value is lesser that the remaining, for same peak value.

Correct Answer: A

Solved Example: 20-2-02

The average value of the periodic waveform shown in the following figure is:

A. 1.414 V

B. 2 V

C. 2.4 V

D. 2.5 V

The average voltage (or current) of a periodic waveform is defined as the area under the waveform divided by time. In other words, the averaging of all the instantaneous values along time axis with time being one full period, (T). $V_{avg} = \dfrac{\dfrac{1}{2}\left(2 \times 3\right)+ \left(3 \times 3\right) +\left(0 \times 1\right)}{\left(2+3+1\right)} = \dfrac{12}{6} = 2\ V$

Correct Answer: B

Learning Objectives:

• Calculate the root-mean-square value of a waveform.

The effective or RMS value of an alternating quantity is that steady current (DC) which when flowing through a given resistance for a given time produces the same amount of heat produced by the alternating current flowing through the same resistance for the same time.

\[X_{\mathrm{eff}} =X_{\mathrm{RMS}}=\left [ \frac{1}{T}\int_{0}^{T}x^{2}(t)dt \right ]^{\frac{1}{2}}\] RMS value of a sinusoidal current =

\[\dfrac{I_{m}}{\sqrt{2}} = 0.707 I_{m}\]

Solved Example: 20-3-01

The R.M.S. value and mean value is the same in the case of: (SSC JE EE Paper 8- Oct 2020 Morning)

A. Triangular wave

B. Sine wave

C. Square wave

D. Half wave rectified sine wave

In case of DC, mean (average), RMS and instantaneous values are same.
In case of symmetrical square wave (square wave with equal positive and negative peaks) also, RMS value = mean value = peak value.

Correct Answer: C

Solved Example: 20-3-02

For the same peak value which of the following wave will have the highest R.M.S. value ? (SSC JE EE Paper 4- Jan 2018 Morning)

A. Square wave

B. Half wave rectified sine wave

C. Triangular wave

D. Sine wave

To get a high RMS value, you want a waveform that is as high as possible most of the time, which happens in case of square wave as the instantaneous value is constant (and high) for half cycle. Since the values are squared while calculating RMS values, during negative half also, same effect takes place.

Correct Answer: A

Solved Example: 20-3-03

For a sine wave with peak value I$_{max}$ the R.M.S. value is:

A. 0.5 I$_{max}$

B. 0.707 I$_{max}$

C. 0.9 I$_{max}$

D. 1.414 I$_{max}$

\begin{align*} i_{RMS} &= \left[\dfrac{1}{\pi}\int_{0}^{\pi}i^{2}(t)dt \right ]^{\frac{1}{2}}\\ &=\left[\dfrac{I_{max}^2}{\pi}\int_{0}^{\pi} \sin ^2 t\ dt \right ]^{\frac{1}{2}}\\ &= \dfrac{I_{max}}{\sqrt{2}\sqrt{\pi}}\left[\int_{0}^{\pi} (1- \cos 2t)\ dt \right ]^{\frac{1}{2}}\\ &= \dfrac{I_{max}}{\sqrt{2}\sqrt{\pi}}\left[ \left(t- \dfrac{\sin 2t}{2}\right)_0^\pi \right ]^{\frac{1}{2}}\\ &= \dfrac{I_{max}}{\sqrt{2}\sqrt{\pi}}\left[ (\pi - 0)-(0-0)\right]^{\frac{1}{2}}\\ &= \dfrac{I_{max}\sqrt{\pi}}{\sqrt{2}\sqrt{\pi}} = \dfrac{I_{max}}{\sqrt{2}} = 0.707I_{max} \end{align*}

Correct Answer: B

Solved Example: 20-3-04

A light bulb is rated at 100W for a 220 V supply. The peak voltage of the source is:

A. 220 V

B. 155.56 V

C. 311 V

D. Cannot be determined

Correct Answer: C

Solved Example: 20-3-05

The RMS value of the periodic waveform shown in the following figure is:

A. 0.96 V

B. 2.35 V

C. 2.5 V

D. 4.9 V

The RMS voltage (or current) of a periodic waveform is defined as the effective value equivalent of a steady DC (constant) value which gives the same effect. \begin{align*} V_{RMS} &= \left[\dfrac{1}{T}\int_{0}^{T}v^{2}(t)dt \right ]^{\frac{1}{2}}\\ &=\left[\frac{1}{6}\int_{0}^{2}v^{2}(t)dt + \int_{2}^{5}v^{2}(t)dt + \int_{5}^{6}v^{2}(t)dt\right]^{\frac{1}{2}}\\ &=\left[\dfrac{1}{6}\int_{0}^{2}(1.5t)^{2}dt + \int_{2}^{5}(3)^{2}dt + \int_{5}^{6}0^{2}dt\right]^{\frac{1}{2}}\\ &= \left[\dfrac{1}{6}\left(6 + 27 + 0\right) \right]^{\frac{1}{2}}\\ &= 2.35\ V \end{align*}

Correct Answer: B