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A resistor connected to a source \(\varepsilon\) of ac voltage as shown in the ciruit digram. The symbol for an ac source in a circuit diagram is \(\Theta\). For simplicity, we consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called ac voltage, be given by

\[\begin{array}{r} V = V_{0}sin\omega t\#(i) \end{array}\]

where \(V_{0}\) is the amplitude of the sinusoidal voltage and \(\omega\) is its angular frequency.

The instantaneous potential drop across the resistor \(R\) is

\[\begin{matrix} & & V_{0}sin\omega t = IR \\ & & \text{~}\text{or}\text{~}\ I = \frac{V_{0}}{R}sin\omega t \\ & I = I_{0}sin\omega t & \text{(ii)} \end{matrix}\]

where \(I\) is the instantaneous current and the current amplitude \(I_{0}\) is given by

\[\begin{array}{r} I_{0} = \frac{V_{0}}{R}\#(iii) \end{array}\]

Equation (iii) is just Ohm's law which for resistors work equally well for both ac and dc voltages. The voltage across a pure resistor and the current through it, given by equation (i) and (ii) are plotted as a function of time in figure. Note, in particular that both V and I reach zero, minimum and maximum values at the same time. Clearly, the voltage and current are in phase for a circuit containing pure resistance.

We see that, like the applied voltage, the current varies sinusoidally and has corresponding positive and negative values during each cycle. Thus, the sum of the instantaneous current values over one complete cycle is zero, and the average current is zero. The fact that the average current is zero, however, does not mean that the average power is zero and that there is no dissipation of electrical energy. As you know, joule heating is given by \(I^{2}R\) and depends on \(I^{2}\) (which is always positive whether I is positive or negative) and not on I. Thus there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor.
The instantaneous power dissipated in the resistor is

\[\begin{array}{r} P = I^{2}R = I_{0}\ ^{2}R\sin^{2}\omega t\#(iv) \end{array}\]

The average value of Power \(P\) over a cycle is

\[\begin{array}{r} \overline{P} = \frac{1}{2}I_{0}^{2}R = I_{rms}^{2}R\#(v) \end{array}\]

Where the bar over a letter(here, P ) denotes its average value.

To express ac power in the same form as dc power \(\left( P = I^{2}R \right)\), as special value of current is used. It is called, root mean square (rms) or effective current and is denoted by \(I_{\text{ms}\text{~}}\).
Similarly, we define the rms voltage or effective voltage
From equation [iii], we have

\[\begin{array}{r} \begin{matrix} & V_{0} = I_{0}R \\ \text{~}\text{or}\text{~} & \frac{V_{0}}{\sqrt{2}} = \frac{I_{0}}{\sqrt{2}}R \\ \text{~}\text{or}\text{~} & V_{rms} = I_{ms}R \end{matrix}\#(ix) \end{array}\]

In terms of rms values, the equation for power and relation between current and voltage in ac circuits are essentially the same as those for the dc case.
In fact, the \(I_{\text{ms}\text{~}}\) or rms current is the equivalent dc current that would produce the same average power loss as the alternating current. Equation [v] can also be written as
\(\overline{P} = V_{ms}^{2}/R = I_{ms}^{2}R\ \) (since \(V_{ms} = I_{ms}R\) )

Practice Exercise

Q. 1 If \(E_{0}\) represents the peak value of the voltage in an ac circuit. Find the rms voltage.
Q. 2 In an ac circuit, the rms value of current \(I_{ms}\) is related to the peak current \(I_{0}\) by the relation:
Q. 3 The electric current in a circuit is given by \(i = i_{0}t/\tau\) for some time. What is the rms current for the period \(t = 0\) to \(t = \tau\)

Answers

Q. \(1\ E_{0}/\sqrt{2}\)
Q. \(2I_{\text{rms}\text{~}} = \left( I_{0}/\sqrt{2} \right)\)
Q. \(3\ i_{0}/\sqrt{3}\)

In the previous section, we saw that the current through a resistor is in phase with the ac voltage. But this is not so in the case of an inductor, a capacitor or a combination. In order to show phase relationship between voltage and current in an ac circuit, we use the motion of PHASORS. The analysis of an ac circuit is facilitated by the use of a phasor diagram. A phasor is a vector which rotates about the origin with angular speed \(\omega\), as shown in figure. The vertical components of phasors V and I represent the sinusoidally varying quantities V and I . The magnitudes of phasors V and I represent the amplitudes or the peak values \(V_{0}\) and \(I_{0}\) of these oscillating quantities. Figure (a) shows the voltage and their relationship at time \(t_{1}\) i.e., corresponding to the circuit show in figure for the case of an ac source connected to a resistor. The projection of voltage and current phasors on vertical axis, i.e., \(V_{0}sin\omega t\) and \(I_{0}sin\omega t\), respectively represent the instaneous value of voltage and current at that instant. As they rotate with frequency \(\omega\), curves in figure(b) are generated which respresent the sinusoidal variation of voltage and current with time.

(a) A phasor diagram for the circuit in figure
(b) Graph of \(V\) and I versus to

From figure (a) we see that phasors V and I for the case of a resistor are in the same phase. This is so for all times. This means that the phase angle between the voltage and the current is zero.

AC Voltage Applied to an Inductor:

An ac source connected to an inductor as shown in the circuit below. Usually, inductors have appreciable resistance in their windings, but we shall assume that this is ideal inductor(having zero resistance). Thus, the circuit is a purely inductive ac circuit. Let the voltage across the source be \(V = V_{0}sin\omega\) t. Using the loop equation. \(\sum_{}^{}\ \varepsilon(t) = 0\), and since there is no resistor in the circuit.

\[\begin{array}{r} V - L\frac{dI}{dt} = 0\#(x) \end{array}\]

An AC source connected to an inductor

where the second term is the self-induced emf in the inductor; and L is the self-inductance of the coil. Combining equation \(\lbrack i\rbrack\) and \(\lbrack x\rbrack\), we have

\[\begin{array}{r} \frac{dI}{dt} = \frac{V}{\text{ }L} = \frac{V_{0}}{\text{ }L}sin\omega t\#(xi) \\ dI = \frac{V_{0}}{\text{ }L}sin\omega tdt\#(xi) \\ I = - \frac{V_{0}}{\omega\text{ }L}cos(\omega t)\#(xi) \end{array}\]

Using \(- cos(\omega t) = sin\left( \omega t - \frac{\pi}{2} \right)\), we have

\[\begin{array}{r} i = I_{0}sin\left( \omega t - \frac{\pi}{2} \right)\#(xii) \end{array}\]

where \(I_{0} = \frac{V_{0}}{\omega\text{ }L}\) is the amplitude of the current. The quantity \(\omega L\) is analogous to the resistance and is called inductive reactance, denoted by \(\mathbf{X}_{L}\) :

\[\begin{array}{r} X_{L} = \omega L = 2\pi fL\#(xiii) \end{array}\]

The dimension of inductive reactance is the same as that of resistance and its SI unit is ohm ( \(\Omega\) ). The inductive reactance limits the current in a pure inductive circuit in the same way as does the resistance in a pure resistive circuit. The inductive reactance is directly proportional to the inductance frequency of the voltage source.

A comparison of equation (i) and (ii) for the source voltage and the current in an inductor shows that the current lags the voltage by \(\frac{\pi}{2}\) or one-quarter \(\left( \frac{1}{4} \right)\) cycle. Figure a shows the voltage and the current phasors in the present case at instant \(t\). The current phasor \(I\) is \(\frac{\pi}{2}\) behind the voltage phasor \(V\). When rotated with frequency \(\omega\) counter-clockwise, they generate the voltage and current given by equation [i] and [xii], respectively and as shown in figure (b)

(a) A phasor diagram for the circuit in figure
(b) Graph of \(V\) and I versus ot.

We see that the current reaches its maximum value later than the voltage by one-fourth of a period \(\left\lbrack \frac{T}{4} = \frac{\pi/2}{\omega} \right\rbrack\). You have seen that an inductor has reactance that limits current similar to resistance in a dc circuit. Does it also consume power like a resistance? Let us try to find out. The instantaneous power supplied to the inductor is

\[\begin{array}{r} \text{(}\ldots\lbrack\text{~}\text{xiv}\text{~}\rbrack\text{)}P_{L} = IV = I_{m}sin\left( \omega t - \frac{\pi}{2} \right)V_{0}sin(\omega t) = - I_{0}V_{0}cos(\omega t) \cdot sin(\omega t) = - - \frac{I_{0}V_{0}}{2}sin(2\omega t) \end{array}\]

So, the average power over a complete cycle is zero
since the average of \(sin(2\omega t)\) over a complete cycle is zero.
Thus, the average power supplied to an inductor over one complete cycle is zero.
Physically, this result means the follows. During the first quarter of each current cycle, the flux through the inductor builds up and sets up a magnetic field and energy is stored in the inductor. In the next quarter of cycle, as the current decrease, the flux decreases and the stored energy is returned to the source. Thus, in each half cycle, the energy which is withdrawn from the source is returned to it without any dissipation of power.

An ac source \(\varepsilon\) connected to a capacitor only, a purely capacitive ac circuit is as shown.

An ac source connected to a capacitor

When the capacitor is connected to an ac source, as in figure, it limits or regulates the current, but does not completely prevent the flow of charge. The capacitor is alternately charged and discharged as the current reverses each half cycle. Let \(q(t)\) be the charge on the capacitor at any time t . The instantaneous voltage \(V(t)\) across the capacitor is

\[\begin{array}{r} V(t) = \frac{q(t)}{C}\#(xv) \end{array}\]

From the Kirchhoff's loop rule, the voltage across the source and the capacitor are equal,

\[V_{0}sin\omega t = \frac{q}{C}\]

To find the current, we use the relation \(I = \frac{dq}{dt}\)

\[I = \frac{d}{dt}\left( V_{0}Csin\omega t \right) = \omega CV_{0}cos(\omega t)\]

Using the relation, \(cos(\omega t) = sin\left( \omega t + \frac{\pi}{2} \right)\), we have

\[\begin{array}{r} I = I_{0}sin\left( \omega t + \frac{\pi}{2} \right)\#(xvi) \end{array}\]

where the amplitude of the oscillating current is

\[I_{0} = \frac{V_{0}}{(1/\omega C)}\]

Comparing it to \(I_{0} = \frac{V_{0}}{R}\) for a purely resistive circuit, we find that ( \(1/\omega C\) ) plays the role of resistance.
It is called capacitive reactance and is denoted by \(\mathbf{X}_{\mathbf{C}}\),

\[\begin{array}{r} X_{C} = 1/\omega C = 1/2\pi fC\#(xvii) \end{array}\]

So that the amplitude of the current is

\[\begin{array}{r} I_{0} = \frac{V_{0}}{X_{C}}\#(xviii) \end{array}\]

The dimension of capacitive reactance is the same as that of resistance and its SI unit is Ohm( \(\Omega\) ). The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as does the resistance in a purely resistive circuit. But it is inversely proportional to the frequency and the capacitance.

A comparison of equation (xvii) with the equation of source voltage equation (i) shows that the current in a capacitor leads the voltage by \(\pi/2\). Figure shows the phasor diagram at an instant \(t\). Here the current phasor I is \(\frac{\pi}{2}\) rad ahead of the voltage phasor V as they rotate counter clockwise. Figure shows the variation of voltage and current with time. We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period.
The instantaneous power supplied to the capacitor is

\[\begin{matrix} & & P_{C} = IV = I_{0}cos(\omega t) \cdot V_{0}sin(\omega t) \\ & & \ = I_{0}V_{0}cos(\omega t)sin(\omega t) \\ & \ = \frac{I_{0}V_{0}}{2}sin(2\omega t) & \text{(xix)} \end{matrix}\]

So, as in the case of an inductor, the average power
Since average of \(sin2\omega t\) over a complete cycle is zero. As discussed in the case of an inductor, the energy stored by a capacitor in each quarter period is returned to the source in the next quarter period.

Thus, we see that in the case of an inductor. the current lags the voltage by \(90^{\circ}\) and in the case of a capacitor, the current leads the voltage by \(90^{\circ}\).

Figure shows a series LCR circuit connected to an ac source \(\varepsilon\). As usual, we take the voltage of the source to be \(V = V_{0}sin\omega t\).

A series LCR circuit connected to an ac source
If \(q\) is the charge on the capacitor and I the current, at time \(t\), we have, from Kirchhoff's loop rule:

\[\begin{array}{r} L\frac{dI}{dt} + IR + \frac{q}{C} = V\#(xx) \end{array}\]

We want to determine the instantaneous current I and its phase relationship to the applied alternating voltage \(V\). We shall use the technique of phasors to solve equation \(\lbrack xx\rbrack\) to obtain the time - dependence of I.

Phasor-Diagram Solution:

From the circuit shown in figure we see that the resistor, inductor and capacitor are in series. therefore, the ac current in each element is the same, having the same amplitude and phase. Let it be

\[\begin{array}{r} I = I_{0}sin(\omega t + \phi)\#(xxi) \end{array}\]

where \(\phi\) is the phase difference between the voltage across the source and the current in the circuit. On the basis of that we construct a phasor diagram for the present case.
Let I be the phasor representing the current in the circuit as given by equation [xxi]. Further, let \(V_{L},V_{R},V_{C}\), and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. From previous section, we know that \(V_{R}\) is in phase with \(I\), \(V_{c}\) is \(\frac{\pi}{2}\) rad behind \(I\) and \(V_{L}\) is \(\frac{\pi}{2}\) rad ahead of \(I.V_{R},V_{C}\) and \(I\) are shown in figure (a) with appropriate phase-relations.

The length of these phasors or the amplitude of \(V_{R},V_{C}\) and \(V_{L}\) are :

\[\begin{array}{r} V_{R0} = I_{0}R,V_{C0} = I_{0}X_{C},V_{L0} = I_{0}X_{L}\#(xxii) \end{array}\]

The voltage equation ( xx ) for the circuit can be written as

\[\begin{array}{r} V = V_{L} + V_{R} + V_{C}\#(xxiii) \end{array}\]

(a)
(a) Relation between the pha

(b)
(b) Relation between the phasors for the circuit in

This relation is represented in figure (b). Since \({\overrightarrow{V}}_{C0}\) and \({\overrightarrow{V}}_{L0}\) are always along the same line and in opposite direction, they can be combined vectorilly into a single phasor \(\left( {\overrightarrow{V}}_{C} + {\overrightarrow{V}}_{L} \right)\) which has a magnitude \(\left| {\overrightarrow{V}}_{C0} - {\overrightarrow{V}}_{L0} \right|\). Since, \({\overrightarrow{V}}_{0}\) is represented as the hypotenuse of a right- angle triangle

\[\begin{array}{r} \text{(}\ldots\lbrack\text{~}\text{xxiv}\text{~}\rbrack\text{)}V_{0}^{2} = V_{R0}^{2} + \left( V_{C0} - V_{L0} \right)^{2} \end{array}\]

Substituting the values of \(V_{R0},{\text{ }V}_{C0}\) and \(V_{L0}\) from equation [xxii] into the above equation, we have

\[\begin{matrix} & V_{0}^{2} = \left( I_{0}R \right)^{2} + \left( I_{0}X_{C} - I_{0}X_{L} \right)^{2} \\ & \ = I_{0}^{2}\left\lbrack R^{2} + \left( X_{C} - X_{L} \right)^{2} \right\rbrack \end{matrix}\]

or \(\ I_{0} = \frac{V_{0}}{\sqrt{R^{2} + \left( X_{C} - X_{L} \right)^{2}}}\)

By analogy to the resistance in a circuit, we introduce the impedance Z in ac circuit :

\[\begin{array}{r} I_{0} = \frac{V_{0}}{Z}\#(xxv) \end{array}\]

where \(Z = \sqrt{R^{2} + \left( X_{C} - X_{L} \right)^{2}}\)

Since phasor I is always parallel to phasor \(V_{R}\), the phase angle \(\phi\) is the angle between \(V_{R}\) and V and can be determined from figure:

\[tan\phi = \frac{V_{C0} - V_{L0}}{{\text{ }V}_{R0}}\]

from the impedance triangle

\[\begin{array}{r} tan\phi = \frac{X_{C} - X_{L}}{R}\#(xxvii) \end{array}\]

impedance triangle

Equations (xxvi) and (xxvii) are shown in figure. This is called Impedance triangle which is a right triangle with Z as its hypotenuse.

Equation ( \(xxv(\) a \()\) ) gives the amplitude of the current and figure gives the phase angle. With these, equation(xxi) is completely specified.

If \(X_{C} > X_{L},\phi\) is positive and the circuit is capacitive. Consequently, the voltage across the source lags the current.

If \(X_{C} < X_{L},\phi\) is negative and the circuit is inductive.
Consequently, the voltage across the source leads the current.
Figure shows the phasor diagram and variation of \(V\) and \(I\) with \(\omega\) the for the case \(X_{C} > X_{L}\).
Thus, we have obtained the amplitude and phase of current for an LCR series circuit using the technique of phasors. But this method of analyzing ac circuits suffers from certain disadvantages. First, the phasor diagram states nothing about the initial condition. One can take any arbitrary value of \(t\) and draw different phasors which show the relative angle between different phasors. The solution so obtained is called the steady-state solution.

(a) Phasor diagram of \(V\) and I. (b) Graphs of \(V\) and I versus \(\omega\) for a series \(LCR\) circuit where \(X_{C} > X_{V}\),

An interesting characteristic of the series RLC circuit is the phenomenon of resonance. The phenomenon of resonance is common among systems that have a tendency to oscillate at a particular frequency. This frequency is called the system's natural frequency. If such a system is driven by an energy source at a frequency that is near the natural frequency, the amplitude of oscillation is found to be large.

For an RLC circuit driven with voltage of amplitude \(V_{0}\) and frequency \(\omega\), we found that the current amplitude is given by \(I_{0} = \frac{V_{0}}{Z} = \frac{V_{0}}{\sqrt{R^{2} + \left( X_{C} - X_{L} \right)^{2}}}\)
with \(X_{C} = \frac{1}{\omega C}\) and \(X_{L} = \omega L\). So if \(\omega\) is varied, then at a particular frequency \(\omega_{0},X_{C} = X_{L}\), the impendance is minimum. This frequency is called the resonant frequency:
then \(\ \frac{1}{\omega_{0}C} = \omega_{0}\text{ }L\ \) or \(\ \omega_{0} = \frac{1}{\sqrt{LC}}\)

At resonant frequency, the current amplitude is maximum; \(I_{0} = \frac{V_{0}}{R}\)
The variation of \(I_{0}\) with \(\omega\) in a RLC series circuit with \(L = 1.00mH,C = 1.00nF\) for two values of R : (i) \(R = 100\Omega\) and (ii) \(R = 200\Omega\) are shown in the figure. For the source applied \(V_{0} = 100\text{ }V.\omega_{0}\) for this case is \(\left( \frac{1}{\sqrt{LC}} \right) = 1.00 \times 10^{6}rad/s\).

\[\text{~}\text{Variation of}\text{~}L_{\text{, with}\text{~}}\omega\text{~}\text{for two cases: (}\text{i}\text{)}\text{~}R = 100\Omega\text{, (ii)}\text{~}R = 200\Omega\text{,}\text{~}\]

We see that the current amplitude is maximum at the resonant frequency.

We have seen that a voltage \(V = V_{0}sin\omega t\) applied to a series RLC circuit drives a current in the circuit given by \(I = I_{0}sin(\omega t + \phi)\) where

\[I_{0} = \frac{V_{0}}{Z}\text{~}\text{and}\text{~}\phi = \tan^{- 1}\left( \frac{X_{C} - X_{L}}{R} \right)\]

Therefore, the instantaneous power P supplied to the source is

\[\begin{matrix} & & P = VI = \left( V_{0}sin\omega t \right) \times \left\lbrack I_{0}sin(\omega t + \phi) \right\rbrack \\ & \ = \frac{V_{0}I_{0}}{2}\lbrack cos\phi - cos(2\omega t + \phi)\rbrack & \text{(xxix)} \end{matrix}\]

The average power over a cycle is given by the average of the two terms in R.H.S. of equation [xxxvii]. Second term average is zero, therefore,

\[\begin{matrix} & & \overline{P} = \frac{V_{0}I_{0}}{2}cos\phi = \frac{V_{0}}{\sqrt{2}}\frac{I_{0}}{\sqrt{2}}cos\phi \\ & \ = V_{ms}I_{mm}cos\phi & \text{(xxx)} \end{matrix}\]

This can also be written as,

\[\begin{array}{r} \overline{P} = I_{ms}^{2}Zcos\phi\#(xxxi) \end{array}\]

So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle \(\phi\) between them. The quantity \(cos\phi\) is called the power factor. Let us discuss the following cases: from impedence triangle \(cos\phi = \frac{R}{z}\)

Case (I) Resistive circuit:

If the circuit contains only pure \(R\), it is called resistive. In that case \(\phi = 0,cos\phi = 1\). There is maximum power dissipation.

Case (II) Purely Inductive or Capacitive Circuit :

If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is \(\frac{\pi}{2}\). Therefore, \(cos\phi = 0\), and no power is dissipated even though a current is flowing in the circuit. This current is sometimes referred to as wattless current.

Case(III) LCR Series Circuit:

In an LCR series circuit, power dissipated is given by equation [xxxviii] where \(\phi = \tan^{- 1}\frac{X_{C} - X_{L}}{R}\). So, \(\phi\) may be non-zero (except \(\frac{\pi}{2}\) ) in a RL or RC or RCL circuit. Even in such cases, power is dissipated only in the resistor.

Case (IV) Power Dissipated at Resonance in LCR Circuit:

At resonance \(X_{C} - X_{L} = 0\), and \(\phi = 0\). Therefore, \(cos\phi = 1\), and \(P = I_{0}\ ^{2}Z = I_{0}\ ^{2}R\). That is, maximum power is dissipated in a circuit (through \(R\) ) at resonance.

Practice Exercise

Q. 1 An ac circuit having resistance ( R ) and inductance ( L ) connected in series with a source of \(\omega\) angular frequency. Find the power factor.
Q. 2 In an LCR series ac circuit, the voltage across each of the components, L, C and R is 50 V . Find the voltage across the LC combination.
Q. 3 If in a series \(L - C - R\) circuit, the voltage across \(R,L\) and \(C\) are \(V_{R},V_{L}\) and \(V_{C}\) respectively. Find the voltage of applied AC source.
Q. 4 In a LR circuit the A.C. source has voltage 220 V and the potential difference across the inductance is 176 V . Find the potential difference across the resistance.

Answers

Q. \(1\ R/\left( R^{2} + \omega^{2}L^{2} \right)^{1/2}\)
Q. 2 Zero
Q. \(3\sqrt{\left\lbrack \left( {\text{ }V}_{R} \right)^{2} + \left( V_{L} - V_{C} \right)^{2} \right\rbrack}\)
Q. 4132 V

Angular frequency variation of power in LCR series circuit.

\[P = P_{\max}\sqrt{\frac{R^{2}}{\sqrt{\left\lbrack R^{2} + \left( L\omega - \frac{1}{C\omega} \right)^{2} \right\rbrack}}}\]

Graph between \(P\&\omega\) as shown in the figure.

\[\omega_{1} = + \frac{R}{2L} + \left( \omega_{r}^{2} + \frac{R^{2}}{4L^{2}} \right)^{1/2}\text{~}\text{and}\text{~}\omega_{2} = - \frac{R}{2L} + \left( \omega_{r}^{2} + \frac{R^{2}}{4L^{2}} \right)^{1/2}\]

Now, \(\ \omega_{1} - \omega_{2} = \frac{R}{L}\)

or \(\ \left( \omega_{r} + \Delta\omega \right) - \left( \omega_{r} - \Delta\omega \right) = \frac{R}{L}\ \) or \(\ 2\Delta\omega = \frac{R}{L}\).
The frequency interval between half maximum power points is known as band width.
The ratio of resonance frequency and band width is known as quality factor (Q).
\(\therefore\ Q = \frac{\omega_{r}}{2\Delta\omega} = \frac{\omega_{r}L}{R}\).

Q factor is a measure of the sharpness of resonance. Resonance will be sharp if the value of bandwidth \((2\Delta\omega)\) is small. This is of course possible only when the power-frequency curve fall steeply around \(\omega = \omega_{r}\).

Q.1. A circuit consists of a series combination of a 50 mH inductor and a \(20\mu\text{ }F\) capacitor connected to \(220\text{ }V,50\text{ }Hz\) supply. The circuit has negligible ohmic resistance. (a) Find the amplitude and rms value of current. (b) Find the rms values of voltage across inductor and capacitor. (c) What is the average power transferred to (i) the inductor, and (ii) the capacitor? (d) What is the total power absorbed by the circuit
Sol. Given \(L = 50mH = 50 \times 10^{- 3}H,C = 20\mu\text{ }F = 20 \times 10^{- 6}\text{ }F,{\text{ }V}_{\text{ms}\text{~}} = 220\text{ }V,\text{ }V = 50\text{ }Hz\) and \(\omega = 2\pi v = 100\pi{rads}^{- 1}\). Therefore, peak value of voltage is

\[V_{0} = \sqrt{2}{\text{ }V}_{rms} = \sqrt{2} \times 220 = 311\text{ }V\]

The peak value of the current in a series LCR circuit is

\[\begin{array}{r} I_{0} = \frac{V_{0}}{\left\lbrack R^{2}\left( \omega L - \frac{1}{\omega C} \right)^{2} \right\rbrack^{1/2}} = \frac{V_{0}}{Z}\#(1) \end{array}\]

Putting R = 0 we have

\[\begin{array}{r} I_{0} = \frac{V_{0}}{\omega L - \frac{1}{\omega C}}\#(2) \end{array}\]

(a) Substituting the values of \(V_{0},\omega,\text{ }L\) and C in equation (2) and solving we get \(I_{0} = - 2.17\text{ }A\). The amplitude of current is the magntidue of modulus of current. Therefore, amplitude of curernt is \(\left| I_{0} \right| = 2.17\text{ }A\).

\[\therefore\ I_{ms} = \frac{I_{0}}{\sqrt{2}} = \frac{2.17}{\sqrt{2}} = 1.53\text{ }A\]

(b) Voltage drop across inductor is

\[\left( V_{L} \right)_{ms} = I_{rms} \times \omega L = 1.53 \times 100\pi \times 50 \times 10^{- 3} = 23.1\text{ }V\]

Voltage drop across capictor is

\[\left( V_{C} \right)_{rms} = \frac{I_{ms}}{\omega\text{ }L} = \frac{1.53}{100\pi \times 20 \times 10^{- 6}} = 243\text{ }V\]

(c) Since in an inductor the voltage leads the current by a phase angle of \(90^{\circ}\), the power transferred to it is 0 . Thus \(P_{L} = 0\). We know that in a capacitor the current leads the voltage by \(90^{\circ}\). Hence power transfered to capcitor is 0 . Thus \(P_{C} = 0\).
(d) Total power absorbed by the circuit is \(P = P_{L} + P_{C} = 0\)
Q. 2 A bulb is rated \(55\text{ }W/110\text{ }V\). It is to be connected to a \(220\text{ }V/50\text{ }Hz\) with inductor in series. What should be the value of inductance so that bulb gets correct voltage.

Sol. We want \(\frac{V_{\text{app}\text{~}}R}{\sqrt{R^{2} + L^{2}\omega^{2}}} = 110\) or \(\frac{220R}{\sqrt{R^{2} + L^{2}\omega}} = 110\)

\[\begin{matrix} & \text{~}\text{or}\text{~}\ 4R^{2} = R^{2} + L^{2}\omega^{2}\text{~}\text{or}\text{~}L\omega = \sqrt{3}R \\ & L(100\pi) = 220\sqrt{3} \\ & L = \frac{2.2\sqrt{3}}{3.14} = \frac{2.2(1.732)}{3.14} = 1.2H;R = \frac{110 \times 110}{55} = 220\Omega \end{matrix}\]

Q. 3 A circuit draws a power of 550 watt from a source of \(220volt,50\text{ }Hz\). The power factor of the circuit is 0.8 and the current lags in phase behind the potential difference. To make the power factor of the circuit as 1.0 , what capacitance will have to be connected with it.

Sol. As the current lags behind the potential difference, the circuit contains resistance and inductance.
Power, \(P = v_{\text{rms}\text{~}} \times i_{\text{rms}\text{~}} \times cos\phi\)
Here, \(i_{\text{rms}\text{~}} = \frac{V_{\text{rms}\text{~}}}{Z}\), where \(Z = \sqrt{\left\lbrack \left( R^{2} + (\omega L)^{2} \right) \right\rbrack}\)

\[\therefore\ P = \frac{V_{\text{rms}\text{~}}^{2} \times cos\phi}{Z}\text{~}\text{or}\text{~}Z = \frac{v_{\text{rms}\text{~}}^{2} \times cos\phi}{P}\]

So, \(Z = \frac{(220)^{2} \times 0.8}{550} = 70.4ohm\)
Now, power factor \(cos\phi = \frac{R}{Z}\) or \(R = Zcos\phi\)

\[\therefore\ R = 70.4 \times 0.8 = 56.32ohm\]

Further, \(Z^{2} = R^{2} + (\omega L)^{2}\) or \((\omega L) = \sqrt{\left( Z^{2} - R^{2} \right)}\)
Or \(\ \omega L = \sqrt{(70.4)^{2} - (56.32)^{2}} = 42.2ohm\)
When the capacitor is connected in the circuit,

\[Z = \sqrt{\left\lbrack R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2} \right\rbrack}\]

\[\text{~}\text{and}\text{~}cos\phi = \frac{R}{\sqrt{\left\lbrack R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2} \right\rbrack}}\]

when \(cos\phi = 1,\omega\text{ }L = \frac{1}{\omega C}\)

\[\begin{matrix} \therefore\ & C = \frac{1}{\omega(\omega L)} = \frac{1}{2\pi f(\omega L)} \\ & \ = \frac{1}{(2 \times 3.14 \times 50) \times (42.2)} \\ & \ = 75 \times 10^{- 6}\text{ }F = 75\mu\text{ }F \end{matrix}\]

Q. 4 Find the average current in terms of \(I_{0}\) for the waveform shown.

Sol. \(\ I = 2I_{0}\frac{t}{T_{0}};0 < t < \frac{T_{0}}{2}\);

\[\begin{matrix} I & \ = 2I_{0}\left( \frac{t}{{\text{ }T}_{0}} - 1 \right);\frac{T_{0}}{2} < t < T_{0} \\ I_{av} & \ = \frac{2}{\text{ }T}\int_{0}^{T_{0}}\mspace{2mu}\mspace{2mu} Idt = \frac{2}{{\text{ }T}_{0}}\left\lbrack \int_{0}^{T/2}\mspace{2mu}\mspace{2mu}\frac{2I_{0}t}{{\text{ }T}_{0}}dt \right\rbrack \\ & \ = \frac{2}{{\text{ }T}_{0}^{2}}\left\lbrack \frac{2I_{0}{\text{ }T}_{0}^{2}}{2 \times 4} \right\rbrack = \frac{I_{0}}{2} \end{matrix}\]

Q.5 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which \(R = 3\Omega,\text{ }L = 25.48mH\), and \(C = 796\mu\text{ }F\). Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor.
Sol. (a) To find the impedance of the circuit, we first calculate \(X_{L}\) and \(X_{C}\).

\[\begin{matrix} & X_{L} = 2\pi fL \\ & \ = 2 \times 3.14 \times 50 \times 25.48 \times 10^{- 3}\Omega = 8\Omega \\ & X_{C} = \frac{1}{2\pi fC} = \frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{- 6}} = 4\Omega \end{matrix}\]

Therefore,

\[Z = \sqrt{R^{2} + \left( X_{L} - X_{C} \right)^{2}} = \sqrt{3^{2} + (8 - 4)^{2}} = 5\Omega\]

(b) Phase difference, \(\phi = \tan^{- 1}\frac{X_{C} - X_{L}}{R} = \tan^{- 1}\left( \frac{4 - 8}{3} \right) = - {53.1}^{\circ}\) since \(\phi\) is negative, the current in the circuit lags the voltage across the source.
(c) The power dissipated in the circuit is

\[P = I_{ms}^{2}R\]

Now, \(\ I_{\text{rms}\text{~}} = \frac{I_{m}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\left( \frac{283}{5} \right) = 40\text{ }A\)
Therefore, \(P = (40\text{ }A)^{2} \times 3\Omega = 4800\text{ }W\)
(d) Power factor \(= cos\phi = cos{53.1}^{\circ} = 0.6\).
Q. 6 An inductor of \(20 \times 10^{- 3}\) henry, a capacitor \(100\mu\text{ }F\) and a resistor \(50\Omega\) are connected in series across a source of \(emfV = 10sin(314t)\). Find the energy dissipated in the circuit in 20 minutes. If resistance is removed from the circuit and the value of inductance is doubled, then find the variation of current with time in the new circuit.

Sol. Here, time of 1 cycle \(T = \frac{2\pi}{\omega} = \frac{2\pi}{314} = 1/50\text{ }s\). So, we have to calculate the average energy as time \(\gg T\).
Energy consumed in time \(t = \left( V_{rms}I_{rms}cos\phi \right)t\)

\[\begin{matrix} & \ = \frac{I_{0}}{\sqrt{2}} \cdot \frac{V_{0}}{\sqrt{2}} \cdot \frac{R}{Z}t \\ & E = \frac{V_{0}^{2}R}{2Z^{2}}t\ \left( \therefore I_{0} = \frac{V_{0}}{Z} \right) \end{matrix}\]

Now, \(Z = \sqrt{R^{2} + \left( \omega L - \frac{1}{\omega c} \right)^{2}}\)

\[\begin{matrix} & \ = \sqrt{(50)^{2} + \left( 314 \times 20 \times 10^{- 3} - \frac{1}{314 \times 100 \times 10^{- 6}} \right)^{2}} \\ & \ = \sqrt{3153.6} \approx 56ohm \end{matrix}\]

\(\therefore\ \) Energy consumed \(= \frac{10^{2} \times 50 \times 20 \times 60}{2 \times 3153.6}\) joule
When resistance is removed,

\[\begin{matrix} & cos\phi = \frac{R}{Z^{'}} = 0\text{~}\text{or}\text{~}\phi = \pi/2 \\ & Z^{'} = \frac{1}{\omega c} - \omega L^{'} = \frac{1}{314 \times 10^{- 4}} - 314 \times 40 \times 10^{- 3}\Omega \\ & \ = 19.3\Omega \\ & I = \frac{E_{0}}{Z^{'}}sin(\omega t + \phi) \\ & \ = \frac{10}{19.3}sin(314t + \pi/2) \\ & \ = 0.52cos314tAmp \end{matrix}\]

Q. 7 A resistor of \(200\Omega\) and a capacitor of \(15.0\mu\text{ }F\) are connected in series to a \(220\text{ }V,50\text{ }Hz\) ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltage more than the source voltage? If yes, resolve the paradox.
Sol. Given

\[\begin{matrix} & R = 200\Omega,C = 15.0\mu\text{ }F = 15.0 \times 10^{- 6}\text{ }F \\ & {\text{ }V}_{rms} = 220\text{ }V,f = 50\text{ }Hz \end{matrix}\]

(a) In order to calculate the current, we need the impedance of the circuit. It is

\[\begin{matrix} & Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{R^{2} + (2\pi fC)^{- 2}} \\ & \ = \sqrt{(200\Omega)^{2} + \left( 2 \times 3.14 \times 50 \times 15 \times 10^{- 6}\Omega \right)^{- 2}} \\ & \ = \sqrt{(200\Omega)^{2} + (212.3\Omega)^{2}} \\ & \ = 291.5\Omega \end{matrix}\]

Therefore, the current in the circuit is

\[I_{rms} = \frac{V_{rms}}{Z} = \frac{220\text{ }V}{291.5\Omega} = 0.755\text{ }A\]

(b) Since the current is the same throughout the circuit, we have

\[\begin{matrix} & V_{R} = I_{mms}R = (0.755\text{ }A)(200\Omega) = 151\text{ }V \\ & {\text{ }V}_{C} = I_{mms}X_{C} = (0.755\text{ }A)(212.3\Omega) = 160.3\text{ }V \end{matrix}\]

The algebraic sum of the two voltage, \(V_{R}\) and \(V_{C}\) is 311.3 V which is more than the source voltage of 220 V . How to resolve this paradox? As you have learnt in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem:

\[V_{R + C} = \sqrt{V_{R}^{2} + V_{C}^{2}} = 220\text{ }V\]

Thus, if the phase difference between two voltage is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.
Q. 8 An LCR circuit has \(L = 10mH,R = 3\Omega\) and \(C = 1\mu\text{ }F\) connected in series to a source of \(15cos\omega t\). Calculate the current amplitude and the average power dissipated per cycle at a frequency that is \(10\%\) lower than the resonance frequency.
Sol. As here resonance frequency,

\[\omega_{0} = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{- 2} \times 10^{- 6}}} = 10^{4}\frac{rad}{\text{ }s}\]

so, \(\omega = \omega_{0} - \frac{10}{100}\omega_{0} = \frac{9}{10}\omega_{0} = 9 \times 10^{3}\frac{rad}{s}\)
and hence, \(X_{L} = \omega L = 9 \times 10^{3} \times 10^{- 2} = 90\Omega\)

\[X_{C} = \frac{1}{\omega C} = \frac{1}{9 \times 10^{3} \times 10^{- 6}} = 111.11\Omega\]

so, \(\ X = X_{C} - X_{L} = 111.11 - 90 = 21.11\Omega\)
and hence, \(Z = \sqrt{R^{2} + X^{2}} = \sqrt{3^{2} + (21.11)^{2}}\)
i.e., \(\ Z = \sqrt{9 + 445.63} = 21.32\Omega\)
and as here \(E = 15cos\omega\) t, i.e., \(E_{0} = 15\text{ }V\),

\[I_{0} = \frac{E_{0}}{Z} = \frac{15}{21.32} = 0.704\text{ }A\]

The average power dissipated,

\[P_{av} = V_{rms}I_{ms}cos\phi = \left( I_{rms} \times Z \right) \times I_{rms} \times \frac{R}{Z}\]

i.e., \(\ P_{av} = I^{2}\ _{ms}R = \frac{1}{2}I_{0}^{2}R\ \left\lbrack \right.\ \) as \(\left. \ I_{ms} = \frac{I_{0}}{\sqrt{2}} \right\rbrack\)
so, \(\ P_{av} = \frac{1}{2} \times (0.704)^{2} \times 3 = 0.74\text{ }W\)
Now as \(f = \frac{\omega}{2\pi} = \frac{9 \times 10^{3}}{2\pi}\frac{\text{~}\text{cycle}\text{~}}{s}\)
So, \(\ \frac{P_{av}}{\text{~}\text{cycle}\text{~}} = \frac{0.74\text{ }J/s}{\left( 9 \times 10^{3}/2\pi \right)\text{~}\text{cycle}\text{~}/s} = \frac{2\pi \times 0.74}{9 \times 10^{3}}\frac{\text{ }J}{\text{~}\text{cycle}\text{~}}\)
i.e., \(\ \frac{P_{av}}{\text{~}\text{cycle}\text{~}} = 5.16 \times 10^{- 4}\frac{\text{ }J}{\text{~}\text{cycle}\text{~}}\).
Q. 9 A series LCR circuit containing a resistance of \(120\Omega\) has angular resonance frequency \(4 \times 10^{5}rads^{- 1}\). At resonance the voltages across resistance and inductance are 60 V and 40 V respectively. Find the values of L and C . At what frequency the current in the circuit lags the voltage by \(45^{\circ}\) ?

Sol. At resonance as \(X = 0,I = \frac{V}{R} = \frac{60}{120} = \frac{1}{2}\text{ }A\)
and as \(V_{L} = {IX}_{L} = I\omega L\ L = \frac{V_{L}}{I\omega}\)
so, \(\ L = \frac{40}{(1/2) \times 4 \times 10^{5}} = 0.2mH\)
and as \(\omega_{0} = \frac{1}{\sqrt{LC}},\ C = \frac{1}{\text{ }L\omega_{0}^{2}}\)
i.e., \(\ C = \frac{1}{0.2 \times 10^{- 3} \times \left( 4 \times 10^{5} \right)^{2}} = \frac{1}{32}\mu\text{ }F\)
Now in case of series LCR circuit,

\[tan\phi = \frac{X_{L} - X_{C}}{R}\]

so current will lag the applied voltage by \(45^{\circ}\) if,

\[tan45 = \frac{\omega L - \frac{1}{\omega C}}{R}\]

i.e., \(\ 1 \times 120 = \omega \times 2 \times 10^{- 4} - \frac{1}{\omega(1/32) \times 10^{- 6}}\)
i.e., \(\ \omega^{2} - 6 \times 10^{5}\omega - 16 \times 10^{10} = 0\)
i.e., \(\ \omega = \frac{6 \times 10^{5} \pm \sqrt{\left( 6 \times 10^{5} \right)^{2} + 64 \times 10^{10}}}{2}\)
i.e., \(\ \omega = \frac{6 \times 10^{5} + 10 \times 10^{5}}{2} = 8 \times 10^{5}\frac{rad}{s}\).
Q. 10 In a series RC circuit R=500 \(\Omega,C = 2\mu\text{ }F,\text{ }V = 282sin(377t)\). The power consumed is
(A) 14100 W
(B) 141 W
(C) 10 W
(D) 14.1 W

Sol. \(\ P = \frac{V_{p}^{2}R}{2Z^{2}} = \frac{282 \times 282 \times 500}{2 \times \left( \sqrt{(500)^{2} + \left( \frac{10^{6}}{2 \times 377} \right)^{2}} \right)^{2}} = \frac{282 \times 282 \times 500}{2 \times 1410 \times 1410} = 10\text{ }W\)
Q. 11 The value of \(L,C\) and R in an LCR series circuit are \(4mH,40pF\) and \(100\Omega\) respectively. The quality factor of the circuit is
(A) 10,000
(B) 100
(C) 1000
(D) 10

Sol. \(\ \omega = \frac{1}{\sqrt{LC}}Q = \frac{L\omega}{R} = \frac{L}{R\sqrt{LC}} = \frac{4 \times 10^{- 3}}{100\sqrt{4 \times 10^{- 3} \times 40 \times 10^{- 12}}} = 100\)
Q. 12 In an LR series AC circuit the angular frequency of applied emf is \(2 \times 10^{4}{rads}^{- 1}\) and the value of resistance is \(20\Omega\). The instant at which the value of emf is maximum \(E_{0}\), the value of current is \(\frac{i_{\text{max}\text{~}}}{\sqrt{2}}\).The inductance in the circuit will be
(A) 1 mH
(B) 40 mH
(C) 8 mH
(D) cannot be predicted

Sol. \(\ \phi = \pi/2 - \pi/4 = \pi/4,tan\pi/4 = 1 = \frac{L\omega}{R}\ \therefore L = 1mH\)
Q. 13 The self inductance of the motor of an electric fan is 10 Henry . In order to impart maximum power at 50 Hz it should be connected to a capacitance of
(A) \(3 \times 10^{- 6}\) Farad
(B) \(2 \times 10^{- 6}\) Farad
(C) \(10^{- 6}\) Farad
(D) \(10^{- 4}\) Farad

Sol. For maximum power to be transferred

\[\begin{matrix} & X_{L} = X_{c}\text{~}\text{or}\text{~}L\omega = \frac{1}{C\omega} \\ & \text{~}\text{or}\text{~}C = \frac{1}{L\omega^{2}} = \frac{1}{10 \times (100\pi)^{2}} = 10^{- 6}\text{ }F \end{matrix}\]

Q. 14 A coil of resistance 200 ohms and self inductance 1.0 henry has been connected to an a.c. source of frequency \(200/\pi Hz\). The phase difference between voltage and current is :
(A) \(30^{\circ}\)
(B) \(63^{\circ}\)
(C) \(45^{\circ}\)
(D) \(\ 75^{\circ}\).

Sol. \(\ tan\phi = \frac{X_{L}}{R} = \frac{2\pi fL}{R} = \frac{2\pi(200/\pi)1}{200} = 2\)

\[\therefore\ \phi = 63^{\circ}.\]

Q. 15 The p.d. across an instrument in an a.c. circuit of frequency \(f\) is \(V\) and the current flowing through it is \(I\) such that \(V = 5cos\pi ff(B)\) volt and \(I = 2sin(2\pi ft)\) amp. The power dissipate in the instrument is :
(A) zero
(B) 10 watt
(C) 5 watt
(D) 2.5 watt.

Sol. As \(V = 5cos\pi ft(2) = 5sin(2\pi ft + \pi/2)\)
And \(I = 2sin(2\pi ft)\)
\(\therefore\ \) phase difference between V and I is \(\phi = \frac{\pi}{2}\)
Average power \(P = V_{\text{ms}\text{~}}I_{\text{ms}\text{~}} \times cos\phi = 0\)
Q. 16 In an L-R circuit, the value of L is \((0.4/\pi)\) henry and the value of R is 30 ohm . If in the circuit, an alternating emf of 200 volt at 50 cycles per second is connected, the impedance of the circuit and current will be :
(A) 11.4 ohm, 17.5 ampere
(B) \(30.7ohm,6.5\) ampere
(C) \(40.4ohm,5\) ampere
(D) \(50ohm,4\) ampere.

Sol. Here \(X_{L} = \omega L = 2\pi fL\)

\[\begin{matrix} & = 2\pi \times 50 \times \frac{0.4}{\pi} = 40\Omega \\ & R = 30\Omega \\ \therefore & Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{30^{2} + 40^{2}} = 50\Omega \\ & I_{rms} = \frac{V_{ms}}{Z} = \frac{200}{50} = 4\text{ }A \end{matrix}\]

Q. 17 In an a.c. circuit, V & I are given by
\(V = 100sin(100t)\) volt.
\(I = 100sin(100t + \pi/3)mA\).
The power dissipated in the circuit is :
(A) \(10^{4}\) watt
(B) 10 watt
(C) 2.5 watt
(D) 5 watt.

Sol. \(\ P = V_{rms}I_{rms}cos\phi\)

\[\begin{matrix} & \ = \frac{100}{\sqrt{2}} \cdot \left( \frac{100}{\sqrt{2}} \times 10^{- 3} \right)cos60^{\circ} \\ & \ = \frac{10}{2} \times \frac{1}{2} = 2.5watt \end{matrix}\]

Q. 18 An alternating voltage (in volts) varies with time t (in seconds) as \(V = 200sin(100\pi t)\)
(A) The peak value of the voltage is 200 V
(B) The rms value of the voltage is 220 V
(C) The rms value of the voltage is \(100\sqrt{2}\text{ }V\)
(D) The frequency of the voltage is 50 Hz

Sol. \(\ V = 200sin(100\pi t)\)
Compare this equation with \(V = V_{0}sin\omega t\)

\[\begin{matrix} & V_{0} = 200\text{ }V,\ {\text{ }V}_{rms} = \frac{V_{0}}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \\ & \omega = 50 \end{matrix}\]

Q. 19 A \(50\Omega\) electric heater is connected to \(100\text{ }V,60\text{ }Hz\) ac supply.
(A) The peak value of the voltatge is 100 V
(B) The peak value of the current in the circuit is \(2\sqrt{2}\text{ }A\)
(C) The rms value of the voltage is 100 V
(D) The rms value of the current is 2 A

Sol. \(\ V_{ms} = 100\text{ }V\). Peak value of voltage \(= 100\sqrt{2}\text{ }V\). Peak value of current \(= \frac{100\sqrt{2}}{50} = 2\sqrt{2}\text{ }A\)

\[I_{ms} = \frac{2\sqrt{2}}{\sqrt{2}} = 2\text{ }A\]

Q. \(20\text{ }L,C\) and R respectively represent inductance, capacitance and resistance. Which of the following combinations have the dimensions of frequency?
(A) \(R/L\)
(B) \(1/RC\)
(C) \(R/\sqrt{LC}\)
(D) \(1/\sqrt{LC}\)

Sol. The dimensions of \(\omega L\) and \(1/\omega C\) are the same as those of resistance, where \(\omega = 2\pi v\).
Q. 21 In a series LCR circuit
(A) the voltage \(V_{L}\) across the inductance leads the current in the circuit by a phase angle of \(\pi/2\)
(B) the voltage \(V_{C}\) across the capacitance lags behind the current by a phase angle of \(\pi/2\)
(C) the voltage \(V_{R}\) across the resistance is in phase with the current
(D) the votage across the series combination of \(L,C\) and R is \(V_{\text{max}\text{~}} = V_{L\text{~}\text{max}\text{~}} + V_{C_{\max}} + V_{R\text{~}\text{max}\text{~}}\).

Sol. The voltage across the combination is given by \(V = \left\lbrack \left( V_{R} \right)^{2} + \left( V_{C} - V_{L} \right)^{2} \right\rbrack^{1/2}\).
Q. 22 If \(V = V_{p}sin(\omega t + \pi/3)\) when will the voltage be maximum for the first time?
(A) \(T/6\)
(B) \(T/12\)
(C) \(T/3\)
(D) None of these

Sol. \(\ sin(\omega t + \pi/3) = 1\) or \(\omega t = \pi/6\) or \(\frac{2\pi t}{T} = \pi/6\) or \(t = \frac{T}{12}\) seconds
Q. 23 If \(i = t^{2},0 < t < T\) then rms value of current is
(A) \(\frac{T^{2}}{\sqrt{2}}\)
(B) \(\frac{T^{2}}{2}\)
(C) \(\frac{T^{2}}{\sqrt{5}}\)
(D) None of these

Sol. \(\ i_{rms}\sqrt{\frac{1}{\text{ }T}\int_{0}^{T}\mspace{2mu}\mspace{2mu} i^{2}dt} = \frac{T^{2}}{\sqrt{5}}\)
Q. 24 In the circuit shown in figure, what will be the readings of voltmeter and ammeter?
(A) \(800\text{ }V,2\text{ }A\)
(B) \(220\text{ }V,2.2\text{ }A\)
(C) \(300\text{ }V,2\text{ }A\)
(D) \(100\text{ }V,2\text{ }A\).

Sol. As \(V_{L} = V_{C} = 300\text{ }V\),
and \(\ V = \sqrt{V_{R}^{2} + \left( V_{L} - V_{C} \right)^{2}}\)
\[\therefore\ V_{R} = V = 220\text{ }V\]

Also \(\ I = \frac{V}{R} = \frac{220}{100} = 2.2\text{ }A\).
Q. 25 An AC voltmeter in an LCR circuit reads 30 V across resistance 80 V across inductance and 40 V across capacitance. The value of applied voltage will be
(A) 25 V
(B) 50 V
(C) 70 V
(D) 150 V

Sol. \(\ V = \sqrt{V_{R}^{2} + \left( V_{L} - V_{C} \right)^{2}} = \sqrt{30^{2} + 40^{2}} = 50\text{ }V\)
Q. 26 In an AC circuit, a resistance of \(3\Omega\), an inductance coil of \(4\Omega\) and a condenser of \(8\Omega\) are connected in series with an AC source of 50 V (rms). The average power loss in the circuit will be
(A) 300 W
(B) 600 W
(C) 400 W
(D) 500 W

Sol. \(\ P = \frac{V_{rms}^{2} \times R}{|Z|^{2}} = \frac{50^{2} \times 3}{5^{2}} = 300\text{ }W\)
Q. 27 The current flowing in a coil is 3 A and the power consumed is 108 W . If the a.c. source is of \(120\text{ }V,50\) Hz , the resistance of the circuit is :
(A) \(24\Omega\)
(B) \(10\Omega\)
(C) \(12\Omega\)
(D) \(6\Omega\).

Sol. From \(P = I^{2}R,R = \frac{P}{I^{2}} = \frac{108}{3^{2}} = 12\Omega\)
Q. 28 If the phase difference between voltage and current is \(\pi/6\) and the resistance in the circuit is \(\sqrt{300}\Omega\), then the impedance of the circuit will be
(A) \(40\Omega\)
(B) \(20\Omega\)
(C) \(50\Omega\)
(D) \(13\Omega\)

Sol. \(\ cos\phi = \frac{R}{|Z|}\) or \(\frac{\sqrt{3}}{2} = \frac{\sqrt{300}}{|Z|}\) or \(Z = 20\Omega\)
Q. 29 The rms voltage of the wave form shown is

(A) 10 V
(B) 7 V
(C) 6.37 V
(D) None of these

Sol. \(\ V_{rms} = \sqrt{\frac{1}{\text{ }T}\int_{0}^{T}\mspace{2mu} 10^{2}dt} = 10\text{ }V\)

Question No. 30 to 32

A \(100\Omega\) resistance is connected in series with a 4 H inductor. The voltage across the resistor is, \(V_{R} = (2.0\text{ }V)sin\left( 10^{3}t \right)\).
Q. 30 Find the expression of circuit current
(A) \(\left( 2 \times 10^{- 2}\text{ }A \right)sin\left( 10^{3}t \right)\)
(B) \(\left( 2 \times 10^{- 3}\text{ }A \right)sin\left( 10^{2}t \right)\)
(C) \(\left( 2 \times 10^{- 3}\text{ }A \right)sin\left( 10^{3}t \right)\)
(D) None of these
Q. 31 Find the inductive reactance
(A) \(2 \times 10^{3}ohm\)
(B) \(3 \times 10^{3}ohm\)
(C) \(4 \times 10^{3}ohm\)
(D) \(5 \times 10^{3}ohm\)
Q. 32 Find amplitude of the voltage across the inductor.
(A) 40 V
(B) 60 V
(C) 80 V
(D) 90 V

Sol. 30 to 32
\[30\ i = \frac{V_{R}}{R} = \frac{(2.0\text{ }V)sin\left( 10^{3}t \right)}{100} = \left( 2.0 \times 10^{- 2}\text{ }A \right)sin\left( 10^{3}t \right) \]\(31X_{L} = \omega L = \left( 10^{3} \right) \times (4H) = 4.0 \times 10^{3}\) ohm
32 Amplitude of voltage across inductor,
\(V_{0} = I_{0}X_{L} = \left( 2.0 \times 10^{- 2}\text{ }A \right)\left( 4.0 \times 10^{3} \right.\ \) ohm \() = 80\) volts

Question No. 33 to 35

If various elements, i.e., resistance, capcitance and inductance which are in series and having values \(1000\Omega,1\mu\text{ }F\) and 2.0 H respectively. Given emf as, \(V = 100\sqrt{2}sin1000t\) volts
Q. 33 Voltage across the resistor is
(A) 70.7 Volts
(B) 100 Volts
(C) 141.4 Volts
(D) 270.7 Volts
Q. 34 Voltage across the inductor is
(A) 70.7 Volts
(B) 100 Volts
(C) 141.4 Volts
(D) 270.7 Volts
Q. 35 Voltage across the capacitor is
(A) 70.7 Volts
(B) 100 volts
(C) 141.4 Volts
(D) 270.7 Volts

Sol. 33 to 35
33 rms value of voltage across the source
\(V_{\text{rms}\text{~}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\) Volt
From question, \(\omega = 1000rad/s\)
\[{i_{\text{rms}\text{~}} = \frac{V_{\text{rms}\text{~}}}{|Z|} = \frac{V_{\text{rms}\text{~}}}{\sqrt{R^{2} + \left( X_{L} - X_{C} \right)^{2}}} = \frac{V_{\text{rms}\text{~}}}{\sqrt{R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2}}} }{= \frac{100}{\sqrt{(1000)^{2} + \left( 1000 \times 2 - \frac{1}{1000 \times 1 \times 10^{- 6}} \right)^{2}}} = 0.0707amp }\]Since the curernt will be same every where in the circuit, therefore
P.D. across resistor \(V_{R} = i_{\text{mas}\text{~}}R = 0.0707 \times 1000 = 70.7\) volts
Q. 36 Figure shows a series LCR circuit connected to a variable frequency 200 V source. \(L = 5H\), \(C = 80\mu\text{ }F\) and \(R = 40\Omega\).
P.D. across inductor \(V_{L} = i_{\text{ms}\text{~}}XL = 0.0707 \times 1000 \times 2 = 141.4\) volt
P.D. across capalitor \(V_{c} = i_{\text{mas}\text{~}}X_{c} = 0.0707 \times \frac{1}{1 \times 1000 \times 10^{- 6}} = 70.7\) volts

Column I

(A) The impedance of the circuit at resonance (in ohm)
(B) The current amplitude at resonance (in A)
(C) The rms potential drop across the inductor at resonance (in volt)
(D) The rms potential drop across the resistor at resonance (in V)
Sol.
\[Z = \left\lbrack R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2} \right\rbrack^{1/2} \]\(\omega = \omega_{r} = 1/\sqrt{\text{~}\text{LC}\text{~}}\) (resonance)
\[{Z = R = 40\Omega }{I_{0} = V_{0}/Z = \frac{\sqrt{2}V_{\text{rms}\text{~}}}{R} = 5\sqrt{2}\text{ }A }{I_{rms} = \frac{V_{rms}}{R} = 5\text{ }A }\]The rms potential drop across L is 1250 V
The rms potential across R is \(= I_{ms} \times R = 5 \times 40 = 200\text{ }V\)

Column II

(P) 1250
(Q) 200
(R) 40
(S) \(5\sqrt{2}\)