(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.

\(Here, \ R=40\Omega , \ { E }_{ v }=220V, \ v=50Hz,\)
\((i) \ { I }_{ v }=\frac { { E }_{ v } }{ R } =\frac { 220 }{ 40 } =5.5 \ A\)
\({ I }_{ 0 }=\sqrt { 2 } { I }_{ v }=1.414\times 5.5=7.8 \ A\)
If alternating current is given by
\(I={ I }_{ 0 } \ sin \ \omega t, \ then\)
\({ I }_{ 0 }={ I }_{ 0 } \ sin{ \omega t }; \ sin \ { \omega t }_{ 1 }=1 \ or \ { \omega t }_{ 1 }=\frac { \pi }{ 2 }\)
\(and \ { I }_{ v }=\frac { { { I } }_{ 0 } }{ \sqrt { 2 } } ={ I }_{ 0 }sin{ \ \omega t }_{ 2 }, \ which \ implies\)
\({ \omega t }_{ 2 }=\frac { \pi }{ 2 } +\frac { \pi }{ 4 } \ \ \therefore \omega \left( { t }_{ 2 }-{ t }_{ 1 } \right) =\frac { \pi }{ 2 } +\frac { \pi }{ 4 } -\frac { \pi }{ 2 } =\frac { \pi }{ 4 }\)
\({ t }_{ 2 }-{ t }_{ 1 }=\frac { \pi }{ 4\omega } =\frac { \pi }{ 4\times 2\pi v } =\frac { 1 }{ 8v } =\frac { 1 }{ 8\times 50 } =2.5\times { 10 }^{ -3 }s=2.5 \ ms\)

Energy of photon, \(\mathrm{E}=\mathrm{hv}\)
This implies,\(\mathrm{E}=\mathrm{h} \frac{\mathrm{c}}{\lambda}\)
Where, \(\mathrm{h}=6.62 \times 10^{-34} \mathrm{js}\)
\( \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}\)
If wave length \(\lambda\) is in meter and energy is in joule then, we will divide E by \(1.6 \times 10^{-19}\) to convert into eV (Electron volt).
\(\therefore \mathrm{E}=\frac{\mathrm{hc}}{\lambda \times 1.6 \times 10^{-19}} \mathrm{eV}\)
(1) For y - rays wave length ranges from  to less that \(10^{-14} \mathrm{~m}\)
Therefore,  \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \times 10^3 \mathrm{eV} \approx 10^4 \mathrm{eV}\)
Thus, \( \lambda=10^{-10} \mathrm{~m}, \text { energy }=10^4 \mathrm{eV} \text { and }\) \( \lambda=10^{-14} \mathrm{~m} \text {, energy }=10^8 \mathrm{eV}\)
Energy of y - rays ranges between 10⁴ to \(10^8 \mathrm{eV}\)
(2) For X - rays wave length ranges from \(10^{-8} \mathrm{~m}\) to \(10^{-7} \mathrm{~m}\) For \(\lambda=10^{-8}\)
Therefore, \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-8} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \approx 10^2 \mathrm{eV}\)
\( \lambda=10^{-13} \mathrm{~m} \text {, energy }=10^7 \mathrm{eV}\)
(3) For violet radiation \(\lambda\) ranges from \(4 \times 10^{-7}\) to \(6 \times 10^{-10}\)
Therefore, for \(\lambda=4 \times 10^{-7}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{eV} =3.1 \mathrm{eV} \approx 10^{10} \mathrm{eV}\)
\(\lambda=6 \times 10^{-10} \mathrm{~m} \text {, Energy }=10^3 \mathrm{eV}\)
Energy of ultraviolet radiation vary between \(10^{10}\) to \(10^3 \mathrm{eV}\).
(4) For visible radiations wave length range from \(4 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-7} \mathrm{~m}\)
Therefore,
For \(\lambda=4 \times 10^{-7} \mathrm{~m}\), and Energy \(=10^{10} \mathrm{eV}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{7 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{VV} \)
\(=1.77 \mathrm{eV} \approx 10^{\circ} \mathrm{eV}\)
(5) For infrared radiation $\lambda$ range from \(7 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-14} \mathrm{~m}\)
Therefore, \(\lambda=7 \times 10^{-7} \text {, energy }=10^{\circ} \mathrm{eV}\)
For \(\lambda=7 \times 10^{-4} \text {, energy }=\frac{1}{1000} \text { times }\)
the other order of  \(10^{-3}\)eV
(6) For micro waves $\lambda$ ranges from 1 mm to 0.3 m
For \(\lambda=1 \mathrm{~mm}\) or \(10^{-3}\)
energy is equal to \( \text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-3} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-3} \mathrm{eV} \approx 10^{-3} \mathrm{eV}\)
For \(\lambda=0.3 \mathrm{~m} \text {, Energy }=4.1 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV} \text {. }\)
(7) For Radio waves $\lambda$ ranges from 1 m to few km For $\lambda=1 \mathrm{~m}$
For λ=1m
Energy is equal to
\(=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^0 \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV}\)
Energy for λ of the order of few km≈10⁻⁶eV
The Energy of a photon that a source produces indicates the spacing of relevant energy levels of the source

Given, amplitude of an electromagnetic wave,
E₀ = 120 N/C
Frequency of wave, v = 50 MHz = 50 \(\times\) 10⁶ Hz
(i) Speed of light in vacuum, \(c=\frac{E_0}{B_0}\)
\(\begin{aligned} B_0=\frac{E_0}{c}= & \frac{120}{3 \times 10^8}=40 \times 10^{-8} \end{aligned}\)
= 400 \(\times\) 10^-9 T = 400 nT
Angular frequency of electromagnetic wave,
\(\omega=2 \pi \nu=2 \times 3.14 \times 50 \times 10^6\)
\(\omega=3.14 \times 10^8 \mathrm{rad} / \mathrm{s}\)
Wave number of electromagnetic wave,
\(k=\frac{\omega}{c}=\frac{3.14 \times 10^8}{3 \times 10^8}=1.05 \mathrm{rad} / \mathrm{m}\)
Wavelength of electromagnetic wave,
\(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{50 \times 10^6}=6.00 \mathrm{~m}\)
(ii) Expression of electric field, E = E₀ sin (kx - \(\omega\)t)
E = 120 sin (1.05x - 3.14 \(\times\)10⁸ t)
Expression of magnetic field B,
B = B₀ sin (kx - \(\omega\)t)
E = 120 sin (kx - \(\omega\)t)
B = 4 \(\times\)10^-7 sin (1.05x - 3.14 \(\times\) 10⁸ t)

(a) See Point Number 6 under the heading ..Chapter At A Glance...
(b) (i) Microwaves are used in radar system, in microwave ovens and for satellite communications etc.
(ii) Ultraviolet rays are used for LASIK eye surgery and to kill germs in water purifiers and hospitals etc.
(iii) Infrared rays are used in remote switches.
(iv) Gamma rays are used for radio therapy in cancer patients and for causing certain nuclear reactions.

\(\text { Given } E_{y}=4 \times 10^{5} \cos \left(3.14 \times 10^{8} t-1.57 x\right) \mathrm{N} / \mathrm{C}\)
∵ General equation of electric field is given by
\(E_{y}=E_{0} \cos (\omega t-k x) \mathrm{N} / \mathrm{C}\)
\(\therefore E_{0}=4 \times 10^{5} \mathrm{~N} / \mathrm{C}, \omega=3.14 \times 10^{8} \mathrm{rad} \mathrm{s}^{-1} \)
\(k=1.57 \mathrm{rad} \cdot \mathrm{m}^{-1} \)
\((i) \lambda=\frac{2 \pi}{k}=\frac{2 \times 3.14}{1.57} \mathrm{~m}=4 \mathrm{~m} \)
\((ii) \mathrm{v}=\frac{\omega}{2 \pi}=\frac{3.14 \times 10^{8}}{2 \times 3.14} \mathrm{~Hz}=5 \times 10^{7} \mathrm{~Hz}\)
(iii) The direction of propagation is +x direction.
\( \text { (iv) } v=\frac{\omega}{k}=\frac{3.14 \times 10^{8}}{1.57} \mathrm{~m} / \mathrm{s}=2 \times 10^{8} \mathrm{~m} / \mathrm{s} \)
\(\text { (v) } \mu=\frac{c}{v}=\frac{3 \times 10^{8}}{2 \times 10^{8}}=1.5 \)
\(\text { (vi) } \because \frac{E_{0}}{B_{0}}=c \)
\(\Rightarrow B_{0}=\frac{E_{0}}{c}=\frac{4 \times 10^{5}}{3 \times 10^{8}} \mathrm{~T}=1.33 \times 10^{-3} \mathrm{~T} \)

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.

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.

\(e=?; \ n=200; \ A=0.16{ m }^{ 2 };\)
\({ B }_{ 1 }=0.10 \ Wb{ m }^{ -2 };\ { B }_{ 2 }=0.30\ Wb{ m }^{ -2 }\)
\({ \phi }_{ 2 }-{ \phi }_{ 1 }=NA\left( { B }_{ 2 }-{ B }_{ 1 } \right) =200\times 0.16\left( 0.30-0.10 \right) \)
\({ \phi }_{ 2 }-{ \phi }_{ 1 }=6.4Wb\)
\(dt=0.05s\)
\(e=\frac { d\phi }{ dt } =\frac { { \phi }_{ 2 }-{ \phi }_{ 1 } }{ dt } \ \left[ in\ magnitude \right] \)
\(e=\frac { 6.4 }{ 0.05 } =\frac { 640 }{ 5 } or\ e=128V\)     

In the inductive circuit,
Rms value of current, I = 15.92 A
Rms value of voltage, V = 220 V
Hence, the net power absorbed can be obtained by the relation,
P = VI cos Φ
Where,
Φ = Phase difference between V and I
For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°.
Hence, P = 0 i.e., the net power is zero.
In the capacitive circuit,
Rms value of current, I = 2.49 A
Rms value of voltage, V = 110 V
Hence, the net power absorbed can ve obtained as:
P = VI Cos Φ
For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°.
Hence, P = 0 i.e., the net power is zero.

(a)  \(\lambda =\frac { c }{ v } =\frac { 3\times 10^{ 8 } }{ 2.0\times 10^{ 10 } } \)
\( =1.5\times 10^{ -2 }m\)
\(E=48Vm^{ -1 }\)
(b) \({ B }_{ 0 }=\frac { E_{ 0 } }{ c } =\frac { 48 }{ 3\times { 10 }^{ 8 } }\)
or  \({ B }_{ 0 }=1.6\times 10^{ -7 } \ T\)
(c) Energy density in E field, 
\({ U }_{ E }=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 }\)
Energy density in B field, 
\({ U }_{ B }=\frac { 1 }{ 2\mu _{ 0 } } B^{ 2 }\)
Using \(E=cB\) and \(c=\frac { 1 }{ \sqrt { \mu _{ 0 }\varepsilon _{ 0 } } } ,\)
We find \({ U }_{ E }={ U }_{ B }\).

More than 1, because the frequency of ultraviolet rays is greater than that of infrared rays.

\(\because \text { Impedance, } Z=\frac{V_{\mathrm{rms}}}{\sqrt{R^{2}+(2 \pi v L)^{2}}}, \text { here } L=\mu_{0} n^{2} A l\)
When the iron rod is inserted in a solenoid, the inductance increases according to L'  \(=\mu_{r} \mu_{0} n^{2} A l, \text { as }\) μ_r for iron is very large. Therefore, the current in the circuit decreases and bulb will glow dimmer.

\(\text { (b) As } X_C-X_L=\frac{1}{2 \pi \vee C}-2 \pi v L\)

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 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
\(I=\frac{V}{X_{L}}=\frac{220 \mathrm{~V}}{7.85 \Omega}=28 \mathrm{~A}\)

90 degree

(i) Microwaves are used for aircraft navigation, their frequency range is 10⁹ Hz to 10¹² Hz.
(ii) X-rays are used to study crystal structure, their frequency range is 10¹⁶ Hz to 10²⁰ Hz.

(i) (a)
(ii) (a): Greenhouse effect is due to infrared rays.
(iii) (a): Ozone layer absorbs the harmful ultraviolet radiations coming from the sun.
(iv) (a): Ozone layer lies in stratosphere.
(v) (b): The atmosphere of earth is richest in infrared radiation.