Because of the differing size and the amount of doping, the emitter and collector cannot be interchanged.

(i) Since aperture does not affect f therefore the image would still be formed at a distance of 10 cm.
(ii) When power is doubled, focal length is halved. So, image would be formed at a distance of 5 cm.

Power factor will be maximum, when Φ = 0 i.e., \(\tan ^{-1}\left(\frac{X_L-X_C}{R}\right)=0\)
\(\therefore X_L=X_C \Rightarrow L \omega=\frac{1}{C \omega}  \)
\(\therefore \omega=\frac{1}{2 \pi \sqrt{L C}}\)
∴ Current I be max \(I_m=\frac{V_m}{R}\)
Hence power factor of a RLC series circuit becomes maximum, when
(i) \(X_L=X_C\)
(ii) Current \(I_m=\frac{V_m}{R}\) will be maximum
(iii) Frequency \(\omega_r=\frac{1}{2 \pi \sqrt{L C}}\)

When the spectrum obtained from the Sun is examined, it consists of large number of dark lines (line absorption spectrum). These dark lines in the solar spectrum are known as Fraunhofer lines.

\(\overset { \rightarrow }{ B } =\mu _{ ° }(\overset { \rightarrow }{ H+ } \overset { \rightarrow }{ M } )\) 
But from equation (3.33), in vector form,
\(\overset { \rightarrow }{ M } ={ x }_{ m }\overset { \rightarrow }{ H } \)
Hence, \(\overset { \rightarrow }{ B } =\mu _{ ° }({ x }_{ m }+1)\overset { \rightarrow }{ H } \Rightarrow \overset { \rightarrow }{ B } =\mu \overset { \rightarrow }{ H } \)
where, \(\mu =\mu _{ ° }({ x }_{ m }+1)\Rightarrow { x }_{ m }+1=\frac { \mu }{ \mu _{ ° } } =\mu _{ r }\)
\(\Rightarrow { x }_{ m }=\mu _{ r }-1\)

I_d=I= 0.2 A

\(Y_1=\overline{A+B}, y_2=\overline{A+B}=A+B, y=\overline{A+B}\)

For total internal reflection,
sin i > sin c
\(n=\frac{1}{sin \ c}\)
\(sin \ c=\frac{1}{n}\)
\(sin \ i>\frac{1}{n}\)
\(n>\frac{1}{sin \ i}\)
n >\(\sqrt{2}\)
n >1.414 = 1.5

\(\vec { B } =\frac { \beta }{ \sqrt { 3 } } =(\hat { i } +\hat { j } +\hat { k } )T\)
Using an equation,
Lorentz force, \(\vec{F}=Il\hat{j}\times\vec B\)
We can get,
Lorentz force \(F=\sqrt { \frac { 2 }{ { 3 } } } \beta Il\)

\(\omega=250 \mathrm{rad} / \mathrm{s}, C=4 \times 10^{-} \mathrm{F}  \)
\(R=100 \Omega, \quad \mathrm{V}_{\mathrm{R}}=40 \mathrm{~V}  \)
\(\therefore I_{\mathrm{R}}=\frac{V_R}{100}=\frac{40}{100}=0.4 \mathrm{~A}  \)
\(\omega=\frac{1}{\sqrt{L C}}  \)
\(\omega^2=\frac{1}{L C}  \)
\((250)^2=\frac{1}{L \times 4 \times 10^{-6}}  \)
\(L=\frac{1}{4 \times(250)^2 \times 10^{-6}}  \)
\(=\frac{1}{4 \times 250 \times 250 \times 10^{-6}}  \)
\(=\frac{1}{1000 \times 10^{-6} \times 250}  \)
\(=\frac{10^3}{250}=\frac{1000}{250}=4 \mathrm{H}\)
Voltage acnoss L, V_t = IX_L
V_L = l x L x ω
= 0.4 x 4 x 250
0.4 x 1000 = 400 V

a) Information (Baseband or input signal):
i) Information can be in the form of a sound signal like speech, music, pictures, or computer data which is given as input to the input transducer.
b) Input transducer:
i) It converts the information which is in the form of sound, music, pictures or computer data into corresponding electrical signals.
ii) The electrical equivalent of the original information is called the baseband signal.
iii) The best example is the microphone that converts sound energy into electrical energy.
c) Transmitter
i) It feeds the electrical signal from the transducer to the communication channel
ii) It consists of circuits such as amplifier, oscillator, modulator, and power amplifier.
iii) Amplifier: The transducer output is very weak and is amplified by the amplifier.
iv) Oscillator: It generates high-frequency carrier wave (a sinusoidal wave) for long distance transmission into space. As the energy of a wave is proportional to its frequency, the carrier wave has very high energy.
v) Modulator: It superimposes the baseband signal onto the carrier signal and generates the modulated signal.
vi) Power amplifier: It increases the power level of the electrical signal in order to cover a large distance.
d) Transmitting antenna:
i) It radiates the radio signal into space in all directions.
ii) It travels in the form of electromagnetic waves with the speed of light.
e) Communication channel:
Communication channel is used to carry the electrical signal from transmitter to receiver with less noise or distortion.
Example: Wires, cables, optical fibres in wireline communication and free space in wireless communication.
f) Receiver:
i) The signals that are transmitted through the communication medium are received with the help of a receiving antenna and are fed into the receiver.
ii) The receiver consists of electronic circuits like demodulator, amplifier, detector etc. The demodulator extracts the baseband signal from the carrier signal.
iii) Then the baseband signal is detected and amplified using amplifiers.
iv) Finally, it is fed to the output transducer.
g) Repeaters:
i) Repeaters are used to increase the range or distance through which the signals are sent.
ii) It is a combination of transmitter and receiver.
iii) The signals are received, amplified, and retransmitted with a carrier signal of different frequency to the destination.
iv) The best example is the communication satellite in space
h) Output transducer:
i) It converts the electrical signal back to its original form such as sound, music, pictures or data.
ii) Examples of output transducers are loudspeakers, picture tubes, computer monitor, etc

Working: The loop PQRS is stationary and is perpendicular to the plane of the paper. When field windings are excited, magnetic field is produced around it. Let the field magnet be rotated in clockwise direction by some external means (Figure). 
(i) Assume that initial position of the field magnet is horizontal. At that instant, the direction of magnetic field is perpendicular to the plane of the loop PQRS. The induced emf is zero. This is represented by origin O in the graph between induced emf and time angle.
(ii) When field magnet rotates through 90°, magnetic field becomes parallel to PQRS. The induced emfs across PQ and RS would become maximum. Since they are connected in series, emfs are added up and the direction of total induced emf is given by Flemin's right hand rule.

(iii) Care has to be taken while applying this rule, the thumb indicates the direction of the motion of the conductor with respect to field. For clockwise rotating poles, the conductor appears to be rotating anticlockwise. Hence, thumb should point to the left. The direction of the induced emf is at right angles to the plane of the paper. For PQ, it is inwards and for RS outwards. Therefore, the current flows along PQRS. The point A in the graph represents this maximum emf.
(iv) For the rotation of 180° from the initial position, the field is again perpendicular to PQRS and the induced emf becomes zero. This is represented by point B.
(v) The field magnet becomes again parallel to PQRS for 270° rotation of field magnet. The induced emf is maximum but the direction is reversed. Thus the current flows along SRQP. This is represented by point C
(vi) On completion of 360°, the induced emf becomes zero and is represented by the point D. From the graph, it is clear that emf induced in PQRS is alternating in nature.
(vii) Therefore, when field magnet completes one rotation, induced emf in PQRS finishes one cycle.

(i) The experimental set up it consists of two metal electrodes which are made of small spherical metals. These are connected to larger spheres and the ends of them are connected to induction coil with very large number of turns.
(ii) This is to produce very high electromotive force (emf). Since the coil is maintained at very high potential, air between the electrodes gets ionized and spark (spark means discharge of electricity) is produced. 
(iii) This discharge of electricity affects another electrode (ring type - not completely closed) which is kept at far distance. This implies that the energy is transmitted from electrode to the receiver (ring electrode) in the form of waves, known as electromagnetic waves. If the receiver is rotated by 90^o then no spark is observed by the receiver.
(iv) This confirms that electromagnetic waves are transverse waves as predicted by Maxwell. Hertz detected radio waves and also computed the speed of radio waves which is equal to the speed of light (3 x 10⁸ ms^-1).

i) The TV signals are frequency modulated. The transmission of TV signals cannot be obtained the ground wave propagation, as such signals get absorbed by ground due to their high frequency.
ii) TV signals cannot be transmitted via skywave propagation as the ionosphere is unable to reflect radio waves of frequencies greater than 40 MHz.
iii) Therefore, the only way out for the transmission of TV signals is that receiving antenna should directly intercept the signal from the transmitting antenna.

β = 120
Base current, \({ I }_{ B }=\frac { 25V }{ 1M\Omega } =\frac { 25 }{ 1\times { 10 }^{ 6 } } =25\mu A\)
We know that
\(\beta =\frac { { I }_{ C } }{ { I }_{ B } } \) (or) 
I_C = β I_B = 120 x 25 μA
= 3000 μA = 3 mA
V_CE = V_CC - I_CR_C
= 25 - (3 mA x 5k) = 10 V 

Current, I = 0.4 A, Magnetic flux, Φ = 4 x 10^-3 Wb, Number of turns, N = 200
Inductance, L = \(\frac { N\Phi }{ I }\)
\(L=\frac { 200\times 4\times 10^{ -3 } }{ 0.4 } =200 \times10^{-3+1}\)
= 200 x 10^-2 = 2H.
∴ Inductance of the coil = 2 H.