The rms current is that value of current which produces the same amount of heat as is produced by the alternating current when flows through the same conductor for the same time period.
Let an instantaneous current I = I_m sin rot is passing through a resistor R. Let it be constant for a very short duration of time dt.
Then, very small amount of heat produced is given by dH = I²R dt
Heat produced for one complete cycle is given by
\(H=\int_{o}^{T} d H=\int_{o}^{T} I^{2} R d t\)
\( \therefore \ H =\int_{0}^{T} I_{m}^{2}\left(\sin ^{2} \omega t\right) R d t=I_{m}^{2} R \int_{0}^{T} \sin ^{2} \omega t d t \)
\(=I_{m}^{2} R \int_{t=0}^{T}\left(\frac{1-\cos 2 \omega t}{2}\right) d t \)
\(H =\frac{I_{m}^{2} R}{2}\left[\int_{0}^{T} d t-\int_{0}^{T}(\cos 2 \omega t) d t\right] \)
\(=\frac{I_{m}^{2} R}{2}\left\{T-\left[\frac{\sin 2 \omega t}{2 \omega}\right]_{0}^{T}\right\} \)
\( \text { As } \omega=2 \pi / T \)
\(\mathrm{H}=\frac{I_{m}^{2} R T}{2}-\frac{I_{m}^{2} R}{2}\left[\sin 2 \times \frac{2 \pi}{T} \times T-\sin 0\right] \)
\((\because \sin 4 \pi=\sin 0=0) \)
\(\therefore \ \mathrm{H}=\frac{I_{m}^{2} R T}{2}\)
\(\)If I_rms is steady current flowing through the same circuit for the same time interval T producing the same amount of heat H, then the heat produced is
\(H=I_{\mathrm{rms}}^{2} R T\)  .......(ii)
Equating (i) and (ii), we get
\( \frac{I_{\mathrm{m}}^{2} R T}{2}=I_{\mathrm{rms}}^{2} R T \)
\(\therefore \frac{I_{\mathrm{m}}}{\sqrt{2}}=I_{\mathrm{rms}} \ \text { or } \ I_{\mathrm{rms}}=0.707 I_{\mathrm{m}} \)

Here, \(X_{ C }=100\Omega ,\ v=50\ Hz\)
\(X'_{ C }=? \ v'=125 \ Hz\)
As \(X_{ C }\propto \frac { 1 }{ v } \therefore \frac { X'_{ C } }{ X_{ C } } =\frac { v }{ v' } =\frac { 50 }{ 125 } =0.4\)
\(X'_{ C }=0.4X_{ C }=40\Omega \)

Here, \(R=40\ \Omega ,\ E_{ v }=200V \ v=50Hz\)
\(I_{ v }=\frac { E_{ v } }{ R } =\frac { 220 }{ 40 } =5.5\ A\)
Max. instantaneous current, \(I_{ v }\)
\(=1.414\times 5.5=7.8\ A\) 

Here, L = 50H, E = 2V, R = 10 ohm, \(\tau =?,\ I_{ 0 }\)
\(\tau =\frac { L }{ R } =\frac { 50 }{ 10 } =5s\)
\(I_{ 0 }=\frac { E }{ R } =\frac { 2 }{ 10 } =0.2A\)

\(Here,L=10 \ mH=10 \times { 10 }^{ -3 }H={ 10 }^{ -2 }H\)
\(I=0.1 \ sin \ 200 \ t\)
\( \frac { dI }{ dt } =0.1 \ cos \ 200t \times 200=20 \ cos \ 200t\)
\({ (\frac { dI }{ dt } ) }_{ ma \times }=20 \times 1\)
\( As \ e=L(dI/dt)\)
\(\\ \therefore { \ \ e }_{ ma \times }=L{ \ (dI/dt) }_{ ma \times }={ 10 }^{ -2 } \times 20=0.2 \ V\)

\( { I }_{ v }=\frac { { E }_{ v } }{ Z } =\frac { 130 }{ 2 } \ \ ...(i)\)
\(Since \ \ Z=\sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } =\sqrt { { \left( 12 \right) }^{ 2 }{ +\left( 2\pi vL \right) }^{ 2 } }\)
\(=\sqrt { { \left( 12 \right) }^{ 2 }+{ \left( 2\pi \times 50\times \frac { 0.05 }{ \pi } \right) }^{ 2 } } \)
\(or \ Z=\sqrt { 144+{ \left( 5 \right) }^{ 2 } } =13\Omega \)
\(So \ \ { I }_{ v } \ =\frac { 130 }{ 13 } =10 \ A\)
\(\\ And \ \ { E }_{ L }={ X }_{ L }I=5\times 10=50 \ V\)

\(Given \ E=220V, \ v=50Hz,\)
\(P=10W, \ voltage \ of \ lamp=30V\)
\(=\frac { 10 }{ 30 } =\frac { 1 }{ 3 } A\)
\(R=\frac { 30 }{ { 1 }/{ 3 } } =90\Omega\)
\(Z=\frac { E }{ I } \)
\(=\frac { 220 }{ { 1 }/{ 3 } } =600\Omega\)
\( Z=\sqrt { { R }^{ 2 }+\frac { 1 }{ { \omega }^{ 2 }{ C }^{ 2 } } }\)
\(=\sqrt { { R }^{ 2 }+\frac { 1 }{ 4{ \pi }^{ 2 }{ v }^{ 2 }{ C }^{ 2 } } }\)
\(or \ \ 600=\sqrt { { (90) }^{ 2 }+\frac { 1 }{ { C }^{ 2 }\times 4\times { \pi }^{ 2 }\times { 50 }^{ 2 } } }\)
\( \\ or \ C=4.866\times { 10 }^{ -6 }F\)
\(=4.866\mu F\)

Given  \(\phi =5{ t }^{ 2 }+7t+9,t=1s=1000ms\)
\(\therefore \)  magnitude of induced e.m.f.
\(e=\left| e \right| =\frac { d\phi }{ dt } =\frac { d }{ dt } \left( 5{ t }^{ 2 }+7t+9 \right) \)
\(=10t+7\)
\(=10\times 1000+7=10007V.\)

\(Given \ C=60\mu F=60\times { 10 }^{ -6 }F\)
\( { \ E }_{ v }=110V,v=60Hz\)
\(Since \ { I }_{ v }=\frac { { E }_{ v } }{ { X }_{ C } }\)
\(\therefore \ { I }_{ v }=\omega C{ E }_{ v }=2\pi v \ C \ { E }_{ v }\)
\( =2\times 3.142\times 60\times 60\times 60\times { 10 }^{ -6 }\times 110\)
\( =2.49A\)
\(or \ { I }_{ v }=2.49A\)

When we start the motor, there is no back emf as the motor is at rest. So, a large current flows through the coil. As the motor rotates, the back emf increases and intake of current decreases.

Conversion of ac of low voltage into ac of high voltage & vice versa. Mutual induction : When alternating voltage is applied to primary windings, emf is induced in the secondary windings.

Energy losses :
(a) Leakage of magnetic flux
(b) Eddy currents
(c) Hysteresis loss
(d) Copper loss

\(\begin{aligned} & R=10 \Omega \\ & L=\left(\frac{100}{\pi}\right) \mathrm{mH}=\frac{100}{\pi} \times 10^{-3} \mathrm{H} \\ & V=141 \sin (100 \pi t) \end{aligned}\)
\( \therefore V_0 =141 \mathrm{~V} \)
\(\therefore \omega t =100 \pi t \)
\( \Rightarrow 2 \pi f =100 \pi \)
\(\Rightarrow f =50 \mathrm{~Hz} \)
\(\text { (i) } Z =\sqrt{\left(X_L\right)^2+R^2=\sqrt{(10)^2+(2 \pi / L)^2}} \)
\(=\sqrt{100+\left(2 \times \pi \times 50 \times \frac{100}{\pi} \times 10^{-3}\right)^2} \)
\(=\sqrt{100+\left(100 \times 100 \times 10^{-3}\right)^2} =\sqrt{100+\left(10^4 \times 10^{-3}\right)^2} \)
\(\begin{aligned} =\sqrt{100+(10)^2} & \equiv \sqrt{200}=10 \sqrt{2} \Omega \end{aligned}\)
\(\text { (ii) } \therefore \tan \phi =\frac{\omega L}{R}=\frac{2 \times \pi \times 50 \times \frac{100}{\pi} \times 10^{-3}}{10} \)
\(=\frac{100 \times 100 \times 10^{-3}}{10}=1 \)
\( \Rightarrow \phi =45^{\circ} \)
\(\text { (iii) } V_L =I X_L=\frac{V_{\text {ras }}}{Z} \cdot \omega L \)
\(=\frac{14 \mathrm{I}}{\sqrt{2} \times 10 \sqrt{2}} \times 100 \pi \times \frac{100}{\pi} \times 10^{-3}=705 \mathrm{volt} \)

\( \text { Given: } L=100 \times 10^{-3} \mathrm{H}, R=100 \Omega, X_{C}=200 \Omega, V_{\mathrm{rms}}=15 \sqrt{2} \mathrm{~V} \)
\( v =\frac{500}{\pi} \mathrm{Hz} \)
\(\therefore \quad X_{L} =2 \pi v L \)
\(=2 \pi \times \frac{500}{\pi} \times 100 \times 10^{-3}=100 \Omega\)
The impedance of the circuit is given by
\( Z =\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}} \)
\(=\sqrt{(100)^{2}+(200-100)^{2}} \)
\(Z =\sqrt{2 \times 10^{4}}=\sqrt{2} \times 100 \)
\(\text { and } \ I_{\mathrm{rms}} =\frac{V_{\mathrm{rms}}}{Z}=\frac{15 \sqrt{2}}{\sqrt{2} \times 100}=0.15 \mathrm{~A} \)
Power dissipated is given by
\( P =\left(I_{\mathrm{rms}}\right)^{2} \times R \)
\( =(0.15)^{2} \times 100=2.25 \mathrm{~W} \)

Given, inductance, L = 0.01 H
Resistance, R = 1 \(\Omega\)
Voltage, V = 200 V
Frequency, v = 50 Hz
Impedance of the circuit, Z \(=\sqrt{R^{2}+X_{L}^{2}}\)
\(=\sqrt{R^{2}+(2 \pi V L)^{2}}=\sqrt{1^{2}+(2 \times 3.14 \times 50 \times 0.01)^{2}}\)
\(=\sqrt{10.86}=3.3 \Omega\)
\(\tan \phi=\frac{\omega L}{R}=\frac{2 \pi v L}{R}=\frac{2 \times 3.14 \times 50 \times 0.01}{1}=3.14\)
\(\Rightarrow \quad \phi=\tan ^{-1}(3.14) \approx 72^{\circ}\)
Phase difference, \(\phi=\frac{72 \times \pi}{180} \mathrm{rad}\) 
Time lag between maximum alternating voltage and current,
\(\Delta t=\frac{\phi}{\omega}=\frac{72 \pi}{180 \times 2 \pi \times 50}=\frac{1}{250} \mathrm{~s}\)