Given that the Inductance of the inductor in the circuit is, L = 5.0 H
Given that the Capacitance of the capacitor in the circuit is , C = 80 μH = 80 x 10 ^- 6 F
Given that Resistance of the resistor in the circuit, R = 40 Ω
Value of Potential of the variable voltage supply, V = 230 V
(a) We know that the Resonance angular frequency can be obtained by the following relation :
\(\omega_{r}=\frac{1}{\sqrt{L C}} \omega_{r}=\frac{1}{\sqrt{5 x 80 x 10-6}} \omega_{r}=\frac{10^{3}}{20}=50 \mathrm{rad} / \mathrm{sec}\)
Thus, the circuit encounters resonance at a frequency of 50 rad/s.
(b) We know that the Impedance of the circuit can be calculated by the following relation :
\(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)
At resonant condition,
X _L = X _C
Z = R = 40 Ω
At resonating frequency amplitude of the current can be given by the following relation :
\(I_{0}=\frac{V_{0}}{Z}\)
where,
V ₀ = peak voltage = \(\sqrt{2} V\)
Therefore,
\(I_{0}=\frac{\sqrt{2 V}}{Z}=\frac{\sqrt{2} \times 230}{40}=8.13 \mathrm{~A}\)
Thus, at resonant condition, the impedance of the circuit is calculated to be 40 Ω and the amplitude of the current is found to be  8.13 A
c) rms potential drop across the inductor in the circuit,
( V _L ) _rms = I x ω _r L
Where,
\(I_{r m s}=\frac{I_{0}}{\sqrt{2}}=\frac{\sqrt{2} V}{\sqrt{2} Z}=\frac{230}{40}=\frac{23}{4} A\)
Therefore, ( V _L ) _rms
\(\frac{23}{4} \times 50 \times 5=1437.5 \mathrm{~V}\)
We know that the Potential drop across the capacitor can be calculated with the following relation :
\(\left(V_{c}\right)_{r m s}=I \times \frac{1}{\omega_{r} C}=\frac{23}{4} \times \frac{1}{50 \times 80 \times 10^{-6}}=1437.5 V\)
We know that the Potential drop across the resistor can be calculated with the following relation :
\(\left(V_{R}\right)_{r m s}=I R=\frac{23}{4} \times 40=230 \mathrm{~V}\)
Now, Potential drop across the LC connection can be obtained by the following relation :
V _{L C}  = I ( X _L − X _C )
At resonant condition,
X _L = X _C
V _{L C} = 0
Therefore, it has been proved from the above equation that the potential drop across the LC connection is equal to zero at a frequency at which resonance occurs.

The supply frequency and the natural frequency are equal at resonance condition in the circuit.
Given Resistance of the resistor, R = 20 Ω
Given Inductance of the inductor, L = 1.5 H
Given Capacitance of the capacitor , C = 35 μF = 30 x 10 ^- 6 F
An AC source with a voltage of V = 200 V is connected to the LCR circuit,
We know that the Impedance of the above combination can be calculated by the following relation,
\(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)
At resonant condition in the circuit , X _L = X _C
Therefore , Z = R = 20 Ω
We know that Current in the network is given by the relation :
\(I=\frac{V}{Z}=\frac{200}{20}=10 A\)
Therefore, the average power that is being transferred to the circuit in one full cycle :
V I = 200 x 10 = 2000 W

a) Given: The peak voltage of supply is 100 V.
The rms voltage is give as,
v m = 2 ×V
Where, the peak value of supply voltage is v m and its rms value is V.
By substituting the given values in the above equation, we get
300= 2 ×V V= 300 2 =212.1 V
Thus, the value of rms voltage is 212.1V.
b) Given: The rms current in an ac circuit is 10 A.
The peak current in the circuit is given as,i m = 2 ×I
Where, the peak current in an ac circuit is i m and its rms value is I.
By substituting the given values in the above equation, we get
i m = 2 ×10 =14.1 A
Thus, the value of peak current in the given ac circuit is 14.1 A.

Given: The values of resistor is 100 Ω and the supply voltage is 100 V.

(a)

The RMS current is given as,
I= V R
Where, the supply voltage is V and the value of resistor is R.
By substituting the given values in the above equation, we get,
I= 220 100 =2.2 A
Thus, the value of RMS current in the conductor is 2.2 A.
(b) Power consumed over a full cycle is given as,
P=VI
Where, the supply voltage is V and the RMS current is I.
By substituting the given values in the above equation, we get
P=220×2.2 =484 W
Thus, power consumed over a full cycle is 484 W.

(a) We know that P = I V cos\(\phi \) where cos\(\phi \) is the power factor. To supply a given power at a given voltage, if cos\(\phi \) is small, we have to increase current accordingly. But this will lead to large power loss (I²R) in transmission.
(b) Suppose in a circuit, current I lags the voltage by an angle \(\phi \). Then power factor \(\phi \) = R/Z.
We can improve the power factor (tending to 1) by making Z tend to R. Let us understand, with the help of a phasor diagram.

how this can be achieved. Let us resolve I into two components. I_p along the applied voltage V and I_q perpendicular to the applied voltage. I_q as you have learnt in Section 7.7, is called the wattless component since corresponding to this component of current, there is no power loss. I_P is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit.
It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current I_q by an equal leading wattless current I'_q. This can be done by connecting a capacitor of appropriate value in parallel so that I_q and I′_q cancel each other and P is effectively I_p V.

Given
R = 200Ω, C = 15.0μF = 15.0 x 10^-6F
V = 220 V, ν = 50 Hz
(a) In order to calculate the current, we need the impedance of the circuit. It is
\(Z=\sqrt{R^{2}+X_{C}^{2}}=\sqrt{R^{2}+(2 \pi v C)^{-2}}\)
\(=\sqrt{(200 \Omega)^{2}+\left(2 \times 3.14 \times 50 \times 15.0 \times 10^{-6} \mathrm{~F}\right)^{-2}}\)
\(=\sqrt{(200 \Omega)^{2}+(212.3 \Omega)^{2}}\)
= 291.67Ω
Therefore, the current in the circuit is
\(I=\frac{V}{Z}=\frac{220 \mathrm{~V}}{291.5 \Omega}=0.755 \mathrm{~A}\)
(b) Since the current is the same throughout the circuit, we have
\(V_{R}=I R=(0.755 \mathrm{~A})(200 \Omega)=151 \mathrm{~V}\)
\(V_{C}=I X_{C}=(0.755 \mathrm{~A})(212.3 \Omega)=160.3 \mathrm{~V}\)
The algebraic sum of the two voltages, 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 V
Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.

(a) To find the impedance of the circuit, we first calculate \(X_{\mathrm{L}}\) and \(X_{\mathrm{C}}\).
\( X_L=2 \pi v L \)
\(=2 \times 3.14 \times 50 \times 25.48 \times 10^{-3} \Omega=8 \Omega \)
\(X_c=\frac{1}{2 \pi v C}  =\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}}=4 \Omega\)
Therefore.
\(Z  =\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{3^2+(8-4)^2} =5 \Omega\)
\(Z  =\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{3^2+(8-4)^2} =5 \Omega\)
(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
\(P=I^2 R\)
Now, \(I=\frac{i_m}{\sqrt{2}}=\frac{1}{\sqrt{2}}\left(\frac{283}{5}\right)=40 \mathrm{~A}\)
Therefore, \(P=(40 \mathrm{~A})^2 \times 3 \Omega=4800 \mathrm{~W}\)
(d) Power factor \(=\cos \phi=\cos \left(-53.1^{\circ}\right)=0.6\)

The capacitance of the capacitor in the circuit is C = 60 μ F or 60 x 10 ^-6 F
The source voltage is V = 110 V
The frequency of the source is ν = 60 Hz
The angular frequency can be calculated using the following relation,
ω = 2πν
The capacitive reactance in the circuit is calculated as follows:
\(X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi \nu C}=\frac{1}{2 \pi \times 60 \times 60 \times 10^{-6}} \Omega\)
Now, the RMS value of the current is determined as follows:
\(I=\frac{V}{X_{C}}=\frac{220}{2 \pi \times 60 \times 60 \times 10^{-6}}=2.49 A\)
Therefore, the RMS current is 2.49 A.

(a) The frequency at which the resonance occurs is
\(\omega_0 =\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{25.48 \times 10^{-3} \times 796 \times 10^{-6}}} \)
\(=222.1 \mathrm{rad} / \mathrm{s} \)
\(v_r =\frac{\omega_0}{2 \pi}=\frac{221.1}{2 \times 3.14} \mathrm{~Hz}=35.4 \mathrm{~Hz}\)
(b) The impedance Z at resonant condition is equal to the resistance:
\(Z=R=3 \Omega\)
The rms current at resonance is
\(=\frac{V}{Z}=\frac{V}{R}=\left(\frac{283}{\sqrt{2}}\right) \frac{1}{3}=66.7 \mathrm{~A}\)
The power dissipated at resonance is
\(P=I^2 \times R=(66.7)^2 \times 3=13.35 \mathrm{~kW}\)

As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.

The capacitive reactance is
\(X_{C}=\frac{1}{2 \pi v C}=\frac{1}{2 \pi(50 \mathrm{~Hz})\left(15.0 \times 10^{-6} \mathrm{~F}\right)}=212 \Omega\)
The rms current is
\(I=\frac{V}{X_{C}}=\frac{220 \mathrm{~V}}{212 \Omega}=1.04 \mathrm{~A}\)
The peak current is
\(i_{m}=\sqrt{2} I=(1.41)(1.04 A)=1.47 A\)
This current oscillates between +1.47A and -1.47 A, and is ahead of the voltage by π/2.
If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Capacitance, C = 30μF = 30 × 10⁻⁶F
Inductance, L = 27 mH = 27 × 10⁻³ H
Angular frequency is given as:

\({ \omega }_{ r }=\frac { 1 }{ \sqrt { LC } } =\frac { 1 }{ \sqrt { 27\times { 10 }^{ -3 }\times 30\times { 10 }^{ -6 } } } =\frac { { 10 }^{ 4 } }{ 9 } =1.1\times { 10 }^{ 3 }rad/s\)
Hence, the angular frequency of free oscillations of the circuit is 1.11 × 10³ rad/s.

Given, inductance, L = 44 mH = 44 x 10^-3H, V_rms = 220V
Frequency of inductor, V = 50 Hz
Inductive reactance, X_L = 2\(\pi\)VL
= 2 x 3.14 x 50 x 44 x I0^-3 = 13.82 \(\Omega\)
The rms value of current in the circuit,
\(I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_{L}}=\frac{220}{13.82}=15.9 \mathrm{~A}\)
Power absorbed, F = V_rms I_rms cos Φ
For pure inductive circuit, Φ = 90\(\unicode{xb0} \)
\(\therefore\) P = 0
Thus, power spent in one half cycle is retrieved in the other half cycle.

The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.

When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (1/ωC) and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.

The inductive reactance.
\(X_{L}=2 \pi v L=2 \times 3.14 \times 50 \times 25 \times 10^{-3} \Omega\)
= 7.85Ω
The rms current in the circuit is
\(I=\frac{V}{X_{L}}=\frac{220 \mathrm{~V}}{7.85 \Omega}=28 \mathrm{~A}\)