i) a ) R = 30 \(\Omega\)
\(\begin{aligned} & C=\frac{250}{\pi} \times 10^{-6} \mathrm{~F} \\ & V=200 \mathrm{~V} \\ & f=50 \mathrm{~Hz} \end{aligned}\)
\(\begin{aligned} Z & =\sqrt{R^2+\left(X_C\right)^2} \\ & =\sqrt{(30)^2+\frac{1}{\omega C}} \\ & =\sqrt{900+\frac{1 \times 10^6}{2 \times \pi \times 50 \times \frac{250}{\pi}}} \\ & =\sqrt{900+\frac{10^6}{250 \times 100}} \end{aligned}\)

\(\begin{aligned} & =\sqrt{900+\frac{1000000}{25000}} \\ & =\sqrt{900+40}=\sqrt{940}=30.65 \Omega \\ I & =\frac{V}{2}=\frac{200}{3065}=6.52 \mathrm{~A} \end{aligned}\)
b) Voltage drops across the resistor = IR
= 652 x 30 = 195.6V
Voltage drops across the capacitor = I x C
\(\begin{aligned} & =652 \times \frac{1 \times 10^6}{2 \times \pi \times 50 \times \frac{250}{\pi}} \\ & =\frac{652 \times 10^6}{100 \times 250 \times 100} \\ & =\frac{652 \times 10^6}{25 \times 10^5}=\frac{652 \times 10^{6-5}}{25} \\ & =\frac{6520}{25}=2608 \mathrm{~V} \end{aligned}\)
c ) Total voltage drop = 260.8+195.6 =456.4 V .Kirchhoff's voltage law is valid when we consider the instantaneous values of voltage 195.6V and 260.3 V are the rms value of voltage. Hence, when there values are added the resultant will be greater than the source voltage.
ii) a) R= 20\(\Omega\), L=2H, C =50x10^-6 F V= 200V
\(\begin{aligned} f_r & =\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{2 \times 50 \times 10^{-6}}} \\ & =\frac{1}{2 \pi \sqrt{100 \times 10^{-6}}}=\frac{1}{2 \pi \sqrt{10^{-4}}} \\ & =\frac{1}{2 \pi \sqrt{\frac{1}{100} \times \frac{1}{100}}}=\frac{1}{2 \pi \times \frac{1}{100}}=\frac{50}{314} \\ & =1592 \mathrm{~Hz} \sim 16 \mathrm{~Hz} \end{aligned}\)
b) At resonance, the impedance of the circuit is minimum, and the current in maximum.
\(I=V / R=\frac{200}{20}=10 \mathrm{~A}\)
Average power transterred to the circuit in one complete cycle.
\( \begin{aligned} \quad P & =1 / 2 \times V_{\text {rus }} \times l_{\text {nas }} \times \cos (\theta) \\ V_{\text {rus }} & =V / \sqrt{2}=200 / \sqrt{2}=141.4 \mathrm{~V} \\ I_{\text {rus }} & =10 / \sqrt{2}=707 \mathrm{~A} \\ P & =1 / 2 \times 141.4 \times 707=499849=500 \mathrm{~W} \end{aligned} \)
c) Potenial drop across the capacitor \(V_c=\frac{1}{2 \pi f C}\)
At resonance,
\(\begin{aligned} V_C & =\frac{l}{2 \pi f C}=\frac{10 \times 10^6}{2 \times 314 \times 16 \times 50}=\frac{10^7}{314 \times 16}=\frac{10^7}{5024} \\ & =1990.4 \mathrm{~V} \end{aligned}\)

(i) A galvanometer is a sensitive device and can measure up to few microampere. hence may get damaged, if strong current is passed through it.
(ii) A galvanometer has larger resistance than an ammeter. Therefore, when it is connected in series with the circuit, the current in the circuit decreases.

Proton will tend to move along the positive X-axis in the direction of a uniform electric field.

When the frequency of the supply equals the natural frequency of the circuit, resonance occurs.
\({ Z }_{ r }=R=20\Omega \)
\({ I }_{ rms }=\frac { { V }_{ rms } }{ { Z }_{ r } } =\frac { 200 }{ 20\quad } =10A\)
Average power transferred in one cycle,
\({ P }_{ av }={ V }_{ rms }{ I }_{ rms }\cos { \phi } \)
= 200 x 10 x cos 0⁰    [\(\phi ={ 0 }^{ 0 }\)]
= 2000W = 2kW

As we know the mutual inductance M is directly proportional to the relative permeability of space, \(\mu\)  So, it will increase on placing a thin iron sheet between the two coils.

The current drawn form d.c. source will increase. This ia because when the coil is stretched and air gaps are created between successive elements of the spiral coil, magnetic flux will leak through the gaps. According to Lewi's law, emf induced must resists the decrease in flux. This can be done when more current is drawn from d.c. source.

For making permanent magnets, a material having high coercivity is used. Steel is a better choice compared to soft iron.

q = 0.48C

At the transmitting point, the voltage is increased and the corresponding current is decreased by using step-up transformer. Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss. At the receiving point, the voltage is decreased and the current is increased to appropriate and then it is given to consumers.

In any transformer
\(\frac{I_s}{I_p}=\frac{E_p}{E_s}=\frac{N_p}{N_s}=\frac{1}{K}\)
Here \(E_p=220 \mathrm{ volts}, I_s=2 \mathrm{Amp}\)
\( \frac{N_p}{N_s}=\frac{1}{25} \)
\( \therefore I_p=\frac{N_s}{N_p} \times I_s=25 \times 2=50 \)
\(\text { and } E_s=\frac{N_s}{N_p} \times E_p=25 \times 250  =55000 \text { Volts } \)
\( \text { Power output }=E_s \times I_s  =5500 \times 2=11000 \text { watts. }\)

The value of carbon resistance = 3700 Ω ± 10%
or R = 37 x 10² ± 10%
The colour assigned to numbers 3, 7 and 2 are orange, violet and red.
For ± 10 % accuracy, the colour is silver.
Hence, the bands of colour on carbon resistance in sequence are orange, violet, red and silver.

The equivalent network of the given network is shown below

Therefore equivalent capacitance,
C_eq = [2 C series C] II [C series 2C]
\(=2\left[\frac{2 C \times C}{2 C+C}\right]=\frac{4 C}{3}\) 

(i) L = 0.048H
(ii) R = 11ohm
(iii) \(\frac { P }{ P' } =\frac { 500 }{ 1600 } =0.31\)

Power of motor,
P = 1 hp = 746 watt.
Electric energy consumed for running a motor for the month of November,
E = 746 x 5 x 30 Wh
= \(\frac{746 \times 5 \times 30}{1000} \ \ kWh\)
Bill of electricity
= \(\frac{746 \times 5 \times 30}{1000}\times 1.50\)
= Rs. 167.85

\(Here,{ N }_{ 1 }=400, \ { N }_{ 2 }=700\)
\( I_{ 1 }=3.5A \ \phi =300 \times { 10 }^{ -6 }{ Tm }^{ 2 }\)
\({ \phi }_{ 2 }=900 \times { 10 }^{ -6 }{ Tm }^{ 2 }\)
\(M=\frac { { N }_{ 2 }{ \phi }_{ 2 } }{ I_{ 1 } } =\frac { 700 \times 90 \times { 10 }^{ -6 } }{ 3.5 } =1.8 \times { 10 }^{ -2 }H\)
\({ L }_{ 1 }=\frac { { N }_{ 1 }{ \phi }_{ 1 } }{ I_{ 1 } } =\frac { 400 \times 300 \times { 10 }^{ -6 } }{ 3.5 } =3.43 \times { 10 }^{ -2 }H\)
\({ e }_{ 2 }=M(\frac { dI_{ 1 } }{ dt } )=1.8 \times { 10 }^{ -2 } \times (0.5)\)
\(=9 \times { 10 }^{ -3 }V\)

\(8 \times { 10 }^{ -5 }Wb;8 \times { 10 }^{ -3 }Wb\)
\(Here,n=100,L=20mH=20 \times { 10 }^{ -3 }H;\)
\( i=4 \ mA=4 \times { 10 }^{ -3 }A\)
Magnetic flux through the cross section of coil 
\(Here,n=100,L=20mH=20 \times { 10 }^{ -3 }H;\)
\(Li=20 \times { 10 }^{ -3 } \times 4 \times { 10 }^{ -3 }=8 \times { 10 }^{ -5 }Wb\)
\(Total \ magnetic \ flu \times =nLi=100 \times 8 \times { 10 }^{ -5 }\)
\(=8 \times { 10 }^{ -3 }Wb\)

Let us consider a loop ABCD in a uniform magnetic field of strength B. and Let the current through the loop be I.

Consider force  on arm AB and force  on arm CD.

The net torque is given by -

Where A is the area of the loop
Now,
The magnetic moment of the loop.

Therefore

This is the expression for the torque acting on a current-carrying loop placed in a magnetic field.
(b) In a radial magnetic field, two sides of the rectangular coil remain perpendicular and other two ides remain parallel to magnetic field lines in all position of the coil.

(a) Electric flux through an area is the product of magnitude of area and the component of electric field vector normal to it.
\(\phi_{\mathrm{E}}=\Delta S(E \cos \theta)=\vec{E} \cdot \Delta \vec{S}\)
Its SI unit is NC^-1 m² 
(b) Electric field intensity due to a thin infinite plane sheet charge: Consider a thin infinite sheet of charge with uniform surface charge density a. To calculate electric field at a point P distant r from the sheet we imagine a symmetrical Gaussian surface in such a way that the point charge lies on it. Here we assume a cylinder of cross-sectional area A and length 2r with its axis perpendicular to the sheet.

Flux through the curved surface of the cylinder,
\(\phi_{1}=\int \vec{E} \cdot \overrightarrow{d s}=0 \left(\because \theta=90^{\circ}\right)\)
Total flux through plane faces of the cylinder,
\(\phi_{2}=2 \int \vec{E} \cdot \overrightarrow{d s}=2 E A \left(\because \theta=0^{\circ}\right)\)
Net flux through the Gaussian surface is
\(\phi=\phi_{1}+\phi_{2}=2 E A\)
Net charge enclosed by the Gaussian surface is
\(Q=\sigma A\)
According to the Gauss's theorem, \(\phi=\frac{Q}{\varepsilon_{0}}\)
\(\therefore \phi=\frac{\sigma A}{\varepsilon_{0}}\)
From equations (i) and (ii), we get
\(2 E A=\frac{\sigma A}{\varepsilon_{0}} \Rightarrow E=\frac{\sigma}{2 \varepsilon_{0}}\)
(c) For positively charged sheet, the electric field is directed away from the sheet.
For negatively charged sheet, the electric field is directed towards the plane sheet.

Max induced emf = 0.603 V
Average induced emf = 0 V
Max current in the coil = 0.0603 A
Average power loss = 0.018 W
(Power comes from the external rotor)
Radius of the circular coil, r = 8 cm = 0.08 m
Area of the coil, A = πr² = π × (0.08)² m²
Number of turns on the coil, N = 20
Angular speed, ω = 50 rad/s
Magnetic field strength, B = 3 × 10⁻² T
Resistance of the loop, R = 10 Ω
Maximum induced emf is given as:
e = Nω AB
= 20 × 50 × π × (0.08)² × 3 × 10⁻²
= 0.603 V
The maximum emf induced in the coil is 0.603 V.
Over a full cycle, the average emf induced in the coil is zero.
Maximum current is given as:
\(I=\frac{e}{R}\)
\(=\frac{0.603}{10}-0.0603A\)
Average power loss due to joule heating:
\(P=\frac{eI}{2}\)
\(=\frac{0.603\times 0.0603}{2}=0.018W\)
The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent. It must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

Given, q = ± 5 \(\mu\)C = ± 5 x 10^-6 C,
2 I = 5mm = 5 x 10^-3 m
x = OX = OY = 10 cm
= 10 x 10^-2 m
E_x = ?
 and E_y =?
Dipole moment,
p = q x 2l
= 5 x 10^-6C x 5 x 10^-3 m 
= 25 x 10^-9 C-m
(i) Now, find out the electric field at point X on the axial line of dipole
\(E_{y}=\frac{2 p x}{4 \pi \varepsilon_{0}\left(x^{2}-l^{2}\right)^{2}}\)  along BX produced
Since  l << x, therefore
\(E_{X}=\frac{2 p}{4 \pi \varepsilon_{0} x^{3}}\)
\(=\frac{2 \times 25 \times 10^{-9} \times 9 \times 10^{9}}{\left(10 \times 10^{-2}\right)^{3}}\)
= 4.5 x 10⁵ NC^-1
(ii) Now, find out the electric field at point Y on equatorial line of dipole
\(\mathbf{E}_{\gamma}=\frac{\mathbf{p}}{4 \pi \varepsilon_{0}\left(x^{2}+l^{2}\right)^{3 / 2}}\)  along a line parallel to BA 
Since,l << x, therefore \(E_{Y}=\frac{P}{4 \pi \varepsilon_{0} x^{3}}\)
\(\therefore \ E_{Y}=\frac{25 \times 10^{-9} \times 9 \times 10^{9}}{\left(10 \times 10^{-2}\right)^{3}}\)
= 2.25 x10⁵ NC^-1 , along a line parallel to BA.

To draw maximum current from a series L-C-R circuit, the circuit at particular frequency
X_L = X_C.
V = \({{1}\over{2\pi\sqrt{LC}}}={{1}\over{2\times314\sqrt{8\times 2\times {10}^{-6}}}}=39.80\) Hz
This frequency is known as the series resonance frequency.
I₀ = \({ { {E}_{0} }\over{ R } }={ {200 }\over{100 } }=2A\)

Sharpness of resonance It is defined as the ratio of the voltage developed across the inductance (L) or capacitance (C) at resonance to the voltage developed across the resistance (R).
Q = \({ { 1 }\over{ R } }=\sqrt{ { { L }\over{ C } } }\)
It may also be defined as the ratio of resonance angular frequency to the bandwidth of the circuit
Q = \({ { {\omega}_{r} }\over{ 2 \triangle \omega } }\)
Circuit become more selective if the resonance is more sharp, maximum current is more, the circuit is close to resonance for smaller range of \((2\triangle \omega)\) of frequencies. Thus, the tuning of the circuit will be good.

Principle: It works on the principle of mutual induction.
Working: When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links with both primary and secondary windings. Let cp be the flux in each turn in the core at a time due to current in the primary when a voltage Vp is applied to it.
\({ V }_{ s }={ E }_{ s }={ N }_{ s }\frac { d\phi }{ dt } \)
The alternating flux also induces an emf, called back emf in the primary given by
\({ V }_{ p }={ E }_{ p }=-{ N }_{ p }\frac { d\phi }{ dt } \)
But \({ E }_{ p }={ V }_{ p }\)
and \({ E }_{ s }={ V }_{ s }\)
So,
\({ V }_{ s }=-{ N }_{ s }\frac { d\phi }{ dt }\)
\( \\ { V }_{ p }=-{ N }_{ p }\frac { d\phi }{ dt } \)
\(\frac { { V }_{ s } }{ { V }_{ p } } =\frac { { N }_{ s } }{ { N }_{ p } } \)
For a step-up transformer, \(\frac { { V }_{ s } }{ { V }_{ p } } >1\)
So,
\(\frac { { V }_{ s } }{ { V }_{ p } } >1\)
(ii) Sources of energy loss in transformer (any two) Flux leakage / Joule's loss in the resistance of windings / Loss due to eddy currents / Hysteresis loss / Humming loss.
(iii) A step-up transformer steps up the voltage while it steps down the current. So the input and output power remain the same (provided there is no loss). Hence there is no violation of the principle of energy conservation.

Here, the mass of the proton
\(​​m=1.6\times { 10 }^{ -24 }g\)
\(=1.6\times { 10 }^{ -27 }kg\)
Charge on proton
\(e=1.6\times { 10 }^{ -19 }C\)
\({ E }_{ y }=400V/cm=4\times { 10 }^{ 4 }V/m,x=5cm=0.05m\)
\({ v }_{ x }=2.5\times { 10 }^{ 8 }cm/s=2.5\times { 10 }^{ 6 }m/s.\)
\(t=\frac { x }{ { v }_{ x } } =\frac { 0.05 }{ 2.5\times { 10 }^{ 6 } } =2\times { 10 }^{ -8 }s\)
\({ F }_{ y }=e{ E }_{ y }\)
\({ F }_{ y }=(1.6\times { 10 }^{ -19 })\times (4\times { 10 }^{ 4 })N\)
\({ a }_{ y }=\frac { { F }_{ y } }{ m } =\frac { 1.6\times { 10 }^{ -19 }\times 4\times { 10 }^{ 4 } }{ 1.6\times { 10 }^{ -27 } } \)
\({ a }_{ y }=4\times { 10 }^{ 12 }m/{ s }^{ 2 }\)
\({ u }_{ y }=0,{ a }_{ y }=4\times { 10 }^{ 12 }m/{ m }^{ 2 },t=2\times { 10 }^{ -8 }s,y=?\)

Let \(\sigma\) be the surface charge density of capacitor plates.
for mica \(E_1={\sigma\over K_1\epsilon_0}\)
and for fibre \(E_2={\sigma\over K_2\epsilon_o}\)  or \({E_1\over E_2}={K_2\over K_1}\)
As \(E_s=6.4\times 10^6V/m\)
\(E_1={K_2\over K_1}\times E_2={2.5\over 8}\times 6.4\times 10^6=2\times 10^6V/m\)
Maximum voltage or capacitor
\(V=E_1d_1+E_2d_2=2\times 10^3+3.2\times 10^3=5200V\)