Here, I = 50 sin 100\(\pi\)t.
Compare it with I = I₀ sin \(\omega\)t = I₀ sin 2\(\pi\)vt
I₀ = 50A, 2\(\pi\)v = 100, v = 50c/s
Mean value of a.c. over positive half cycle
\(=\frac { 2{ I }_{ 0 } }{ \pi } =\frac { 2\times 50 }{ 3.14 } =31.8A\)
\({ I }_{ v }=\frac { { I }_{ 0 } }{ \sqrt { 2 } } =\frac { 50 }{ 1.414 } =35.35A\)
\(From \ I={ I }_{ 0 }sin \ \omega t\)
\(I=50\ sin\ 2\pi \times 50\times \frac { 1 }{ 600 } =50\times \frac { 1 }{ 2 } =25A\)

\(Here,\ I=5sin\ 100\pi t\)
\( Compare\ with\ I={ I }_{ 0 }sin\ \omega t\)
\({ I }_{ 0 }=5A,\ \omega =2\pi v=100\pi\)
\(v=\frac { 100\pi }{ 2\pi } =50 \ Hz\)
\(\ { I }_{ v }=\frac { { I }_{ 0 } }{ \sqrt { 2 } } =\frac { 5 }{ \sqrt { 2 } } =\frac { 5\sqrt { 2 } }{ 2 } =3.54A\)

\(Here, \ { I }_{ v }=5.0A, \ v=50Hz\)
\((i) \ { I }_{ 0 }=\sqrt { 2 } { I }_{ v }=1.414\times 5.0=7.07A\)
\((ii) \ { I }_{ m }=\frac { 2 }{ \pi } { I }_{ 0 }=\frac { 2 }{ 3.14 } \times 7.07=4.5A\)
\((iii) \ From \ I={ I }_{ 0 } \ sin \ \omega t={ I }_{ 0 }sin2\pi \ vt\)
\(=7.07sin2\pi \times 50\times \frac { 1 }{ 300 } =\frac { 7.07\sqrt { 3 } }{ 2 } =6.12A\)

\(Here, \ C=1.0\mu F={ 10 }^{ -6 }F\)
\(R={ 10 }^{ 4 }ohm, \ { E }_{ 0 }=2.0volt\)
\(During \ charging \ of \ the \ condenser,\)
\(q={ q }_{ 0 }\left( 1-{ e }^{ -t/RC } \right) \ I=-{ I }_{ 0 }{ e }^{ -t/RC }, \ where\)
\({ I }_{ 0 }=\frac { E }{ R } =\frac { 2.0 }{ { 10 }^{ 4 } } =2.0\times { 10 }^{ -4 }amp\)
\(At \ t=0.02s, \ I={ I }_{ 0 }\left( { e }^{ -0.02/{ 10 }^{ 4 }\times { 10 }^{ -6 } } \right) =2\times { 10 }^{ -4 }{ e }^{ -2 }\)
\(=\frac { 2\times { 10 }^{ -4 } }{ { e }^{ 2 } } =\frac { 2\times { 10 }^{ -4 } }{ \left( 2.718 \right) ^{ 2 } } =0.27\times { 10 }^{ -4 }A\)
\(I=27\times { 10 }^{ -6 } \ A=27\mu A\)

\(Here, \ l \ = \ 50cm \ =\frac { 1 }{ 2 } m\)
\(Total \ no. \ of \ turns \ in \ one \ solenoid \ { N }_{ 1 }=20\times 50=1000\)
\(Area \ of \ cross \ section \ of \ outer \ solenoid, \ { A }_{ 1 }=40{ cm }^{ 2 }=40\times { 10 }^{ -4 } \ { m }^{ 2 }\)
\(Total \ no.of \ turns \ in \ inner \ solenoid, \ { N }_{ 2 }=25\times 50=1250\)
\(area \ of \ cross \ section \ of \ inner \ solenoid, \ { A }_{ 2 }=25{ cm }^{ 2 }=25\times { 10 }^{ -4 }{ m }^{ 2 }\)
\(M=\frac { { \mu }_{ 0 }{ N }_{ 1 }{ N }_{ 2 } }{ l } { A }_{ 2 }\)
\(=\frac { 4\pi \times { 10 }^{ -7 }\times 1000\times 1250\times 25\times { 10 }^{ -4 } }{ 1/2 }\)
\(=7.58\times { 10 }^{ -3 }\ henry\)

Given, M = 1.5 H, \(\Delta\)i₁ = 5A, \(\Delta\)t = 10^-3s
We know that, \(M=-\frac{c_2}{\Delta i_1 / \Delta t}\)
\(\Rightarrow \quad e_2=M \times \frac{\Delta i_1}{\Delta t}=1.5 \times \frac{5}{10^{-3}}=7.5 \times 10^3 \mathrm{~V}\)

Here, N = 200, L = 10mH = 10\(\times\)10^-3H,  I = 4mA = 4\(\times\)10^-3 A;\(\phi \) = ?
Total magnetic flux linked with the coil \(\phi \) = NLI = 200\(\times\)(10\(\times\)10^-3)4\(\times\)10^-3 = 8\(\times\)10^-3 Wb.
Magnetic flux through the cross-section of the coil=magnetic flux linked with each turn
= \(\frac { \phi }{ N } =\frac { 8\times { 10 }^{ -3 } }{ 200 } =4\times { 10 }^{ -5 }Wb\)

\(Here, \ dI=?, \ L=10 \ H, \ e=300 \ V,\)
\(dt={ 10 }^{ -2 }sec.,\ d\phi =?\)
\(As \ e=\frac { LdI }{ dt } \ \therefore \ 300=10\frac { dI }{ { 10 }^{ -2 } } \ dI=\frac { 300\times { 10 }^{ -2 } }{ 10 } =0.3A\)
\(Also, \ e=\frac { d\phi }{ dt } \ \ \therefore \ d\phi =edt=300\times { 10 }^{ -2 }=3Wb\)

\(Here,\ \phi =\left( { 5r }^{ 3 }+{ 4r }^{ 2 }+2t-5 \right) \)
\(i=?\ t=2s,\ R=5ohm\)
\(e=\left| \frac { d\phi }{ dt } \right| =\frac { d }{ dt } \left( 5t^{ 3 }+{ 4t }^{ 2 }+2t-5 \right) ={ 5t }^{ 2 }+8t+2={ 15(2)}^{ 2 }+8(2)+2=78V\)
\(i=\frac { e }{ R } =\frac { 78 }{ 5 } =15.6 \ A\)

\(Here,\ { B }_{ 1 }=10T,\ \theta ={ 0 }^{ \circ },\ N=50,\)
\(A=50{ cm }^{ 2 }=50\times { 10 }^{ -4 }{ m }^{ 2 },\ e=?,\)
\({ B }_{ 2 }=0,\ dt=0.1s\)
\(e=\frac { -|{ \phi }_{ 2 }-{ \phi }_{ 1 }| }{ dt } \ =\frac { -\left( 0-{ NB }_{ 1 }A{ cos0 }^{ \circ } \right) }{ dt } \)
\(=\frac { 50\times 10\times 50\times { 10 }^{ -4 }\times 1 }{ 0.1 } \ =25V\)