**Alternating current**

A type of current or voltage whose magnitude changes with time and direction reverse periodically is known as alternating current or alternating voltage.

The instantaneous value of alternating current is

I = I_{o} sin $\omega t$.

Where I_{o} is the maximum (peak) value of current.

The instantaneous value of alternating voltage is

E = E_{o} sin$\omega t$.

Where E_{o} is maximum (peak) value of voltage (emf)

**Measurement of alternating current**

1. The average value of alternating current measured in complete cycle is zero.

2. The average value of alternating current measured in half cycle is known as average value. The average value of alternating current is

I_{Avg.} = 0.637 I_{o}

This value is not in good agreement with practical value.

**3. Root mean square value of current (I _{r.m.s})**

Root Mean Square value of current** **is defined as the steady current (D.C.) which develops heat in a given resistance (R) for a given time (t) as that done by alternating current for the same resistance in same time. It is denoted by I_{rms}.

Let us consider a resistor of resistance R is connected with an AC source. Let I = I_{o} sin$\omega t$ be the instantaneous value of current flowing through the resistance. The small amount of work done by the alternating current in small time dt is

dw = I^{2} Rdt

or, dw = I_{o}^{2} sin^{2} $\omega t$ Rdt

or, dw = I_{o}^{2} R sin^{2$\omega t$} dt ……………… (i)

The total amount of work done in a complete cycle can be obtained by integrating equation (i) from limit O to T.

W =$\int\limits_{0}^{T}{{}}$I_{o}^{2} R sin^{2$\omega t$} dt

or, W = I_{o}^{2} R $\int\limits_{0}^{T}{{}}$ sin^{2$\omega t$} dt

or, W = I_{o}^{2} R $\int\limits_{0}^{T}{{}}$$\left( \frac{1-\cos 2\omega t}{2} \right)$dt

or, W = $\frac{{{I}_{o}}^{2}R}{2}$ $\left( \text{ }\int\limits_{0}^{T}{\text{dt}}-\int\limits_{0}^{T}{\cos 2\omega t\text{ dt}} \right)$

or, W = $\frac{{{I}_{o}}^{2}R}{2}$ [T – O] $\left( \because \text{ }\int\limits_{0}^{T}{\cos 2\omega t\text{ dt =0}} \right)$

or, W =$\frac{{{I}_{o}}^{2}RT}{2}$ …………………….. (ii)

Let I_{rms} be the steady value of current which is passed through the same-resistance ‘R’ for same time ‘t’. Now heat developed by the current is

H = I_{rms}^{2} RT ………………….. (iii)

According to the definition,

H = W

or, I_{rms}^{2} RT = $\frac{{{I}_{o}}^{2}RT}{2}$

or, I_{rms}^{2} = $\frac{{{I}_{o}}^{2}}{2}$

or, I_{rms} =$\frac{{{I}_{o}}}{\sqrt{2}}$

$\therefore $ I_{rms} = 0.7070 I_{o}

Similarly,

E_{rms} = 0.707 E_{o}

This value is in good agreement with practical value.

**Wave diagram and phase diagram.**

(i) E = E_{o} sin$\omega t$ and I = I_{o} sin$\omega t$

(ii) E = E_{o} sin$\omega t$ and I = I_{o} sin$(\omega t+\phi )$

(iii) E = E_{o} sin$\omega t$ and I = I_{o} sin$(\omega t-\phi )$

**AC circuit containing resistor only.**

Let us consider a resistor of resistance R is connected with an AC source of emf E = E_{o} sin$\omega t$ where E is instantaneous value of emf and E_{o} is peak value of emf. The instantaneous value of current flowing through the circuit is

I = $\frac{E}{R}$

Or, I = $\frac{{{E}_{o}}sin\omega t}{R}$

Or, I = I_{o} sin$\omega t$ $\left( \because {{I}_{o}}=\frac{{{E}_{o}}}{R} \right)$

Since we are started from E = E_{o} sin$\omega t$ and arrived at I = I_{o} sin$\omega t$, the wave and phase diagram are:

**AC circuit containing inductor only.**

*Fig: AC circuit containing inductor only.*

Let us consider an inductor of inductance L is connected with an AC source of emf E = E_{o} sin$\omega t$. When the inductor is connected with AC source, changing current flows through it due to which an amount of emf induced in the coil. Let the emf induced in the coil is

$\varepsilon $ = – L $\frac{dI}{dt}$

From Kirchhoff’s second law (KVL),

E + $\varepsilon $ = O

Or, E – L $\frac{dI}{dt}$ = 0

Or, E_{o} sin$\omega t$ = L $\frac{dI}{dt}$

Or, dI = $\frac{{{E}_{o}}}{L}$ sin$\omega t$ dt.

On integrating above equation; we get;

$\int{dI}$ = $\frac{{{E}_{o}}}{L}$$\int{\sin \omega t}$ dt

Or, I =$\frac{{{E}_{o}}}{L}$$\left( -\frac{\cos \omega t}{\omega } \right)$

Or, I = – $\frac{{{E}_{o}}}{\omega L}$ cos$\omega t$

Or, I = – $\frac{{{E}_{o}}}{{{X}_{L}}}$ cos$\omega t$ (Where X_{L} = $\omega L$ is inductive reactance i.e. total resistance offered by inductor.)

Or, I = – I_{o} cos$\omega t$

Or, I = – I_{o} sin$\left( \frac{\pi }{2}-\omega t \right)$

Or, I = I_{o} sin$\left( \omega t-\frac{\pi }{2} \right)$

Since we have started from E = E_{o} sin$\omega t$ and arrived at I = I_{o} sin$\left( \omega t-\frac{\pi }{2} \right)$, the wave and phase diagram are;

**AC circuit containing capacitor only. **

*Fig: AC circuit containing capacitor only. *

Let us consider a capacitor of capacitance C is connected with an AC source of emf E = E_{o} sin$\omega t$. Let V_{c} be the potential across the capacitor. From Kirchhoff’s second law (KVL),

E = V_{c}

Or, V_{c} = E_{o} sin$\omega t$

Let us consider a small amount of charge (dq) flows in small time (dt). Now, the current is

I = $\frac{dq}{dt}$

Or, I = $\frac{d({{V}_{c}}C)}{dt}$ since, q = VC

Or, I = $\frac{d}{dt}$ ( C E_{o} sin$\omega t$ )

Or, I = E_{o} C $\frac{d}{dt}$( sin$\omega t$ )

Or, I = E_{o} C ($\omega \cos \omega t$)

Or, I = E_{o$\omega C$}cos$\omega t$

Or, I = $\frac{{{E}_{o}}}{1/\omega C}$ cos$\omega t$

Or, I = $\frac{E{}_{o}}{{{X}_{L}}}$ cos$\omega t$ (Where X_{c} = $\frac{1}{\omega c}$ is capacitive reactance

i.e. resistance offered by capacitor in a.c. circuit)

Or, I = I_{o} cos$\omega t$

Or, I = I_{o} sin$\left( \omega t+\frac{\pi }{2} \right)$

Since, we have started from E = E_{o} sin$\omega t$ and arrived at I = I_{o} sin$\left( \omega t+\frac{\pi }{2} \right)$, the wave and phase diagram are;

**AC circuit containing resistor and inductor in series. **

*Fig: AC circuit containing resistor and inductor in series.*

Let us consider a resistor of resistance R and an inductor of inductance L are connected in series with an AC source of emf E = E_{o} sin$\omega t$. Let V_{P} and V_{L} be the potential across resistor and inductor respectively.

In AC circuit containing resistor only the current I and voltage V_{R} will in the same phase. The voltage V_{R} is reactant by OA in the figure.

In an AC circuit containing an inductor only, the voltage leads the current by phase angle $\frac{\pi }{2}$ . The voltage V_{L} is represented by OC in the figure.

From $\Delta $OAB,

(OB)^{2} = (OA)^{2} + (AB)^{2}

Or, (OB)^{2} = (OA)^{2} + (OC)^{2}

Or, V_{RL}^{2} = V_{R}^{2} + V_{L}^{2}

Or, V_{RL}^{2} = I^{2}R^{2} + I^{2}X_{L}^{2} ( X_{L} = Inductive reactance)

Or, V_{RL}^{2} = I^{2} (R^{2} + X_{L}^{2})

Or, $\frac{{{V}_{RL}}^{2}}{{{I}^{2}}}$ = R^{2} + X_{L}^{2}

Or, Z^{2} = R^{2} + X_{L}^{2} (where Z = $\frac{{{V}_{RL}}}{I}$ is known as impedance i.e. total resistance offered by a.c. circuit.)

$\therefore $ Z = $\sqrt{{{R}^{2}}+{{X}_{L}}^{2}}$

**Calculation of phase ****angle ($\phi $):**** **

In figure, ∠AOB =$\phi $ is the phase angle.

From $\Delta $OAB

Tan$\phi $ = $\frac{AB}{OA}$ = $\frac{OC}{OA}$

Or, Tan$\phi $ = $\frac{{{V}_{L}}}{{{V}_{R}}}$

Or, Tan$\phi $ = $\frac{I{{X}_{L}}}{IR}$

Or, Tan$\phi $ = $\frac{{{X}_{L}}}{R}$

$\therefore \phi ={{\tan }^{-1}}\left( \frac{{{X}_{L}}}{R} \right)$

The above result shows that voltage leads the current by phase angle$\phi $,

i.e. current lags behind the voltage by phase angle$\phi $.

**AC circuit containing resistor and capacitor in series.**

Let us consider a capacitor of capacitance C and a resistor of resistance R are connected in series with an AC source of emf

E = E_{o} sin$\omega t$. Let V_{R} = IR and V_{C} = IX_{L} be the potential across resistor and capacitor, respectively.

When an AC circuit is connected with a resistor only, the current I and voltage V_{R} will be in the same phase. The voltage V_{R} represented by OA in the figure.

When an AC circuit is connected with a capacitor only, the voltage lags behind the current by phase angle $\frac{\pi }{2}$ . The voltage V_{C} is represented by OC.

Let us complete a parallelogram OABC. Now, the diagonal OB represents effective voltage E (i.e. V_{RC}).

From $\Delta $OAB,

(OB)^{2} = (OA)^{2} + (AB)^{2}

or, (OB)^{2} = (OA)^{2} + (OC)^{2}

or, E^{2} = V_{R}^{2} + V_{C}^{2}

or, E^{2} = (IR)^{2} + (IX_{C})^{2}

or, $\frac{{{E}^{2}}}{{{I}^{2}}}$ = R^{2} + X_{C}^{2}

or, Z^{2} = R^{2} +X_{C}^{2} (where Z = $\frac{E}{I}$ is impedance total resistance offered by a.c. circuit.)

$\therefore $ Z =$\sqrt{{{R}^{2}}+{{X}_{C}}^{2}}$

**Calculation of phase angle:**

From $\Delta $OAB,

Tan$\phi $ = $\frac{AB}{OA}$ = $\frac{OC}{OA}$

or, Tan$\phi $ = $\frac{{{V}_{C}}}{{{V}_{R}}}$

or, Tan$\phi $ = $\frac{I{{X}_{C}}}{IR}$

or, Tan$\phi $ = $\frac{{{X}_{C}}}{R}$

$\therefore \phi ={{\tan }^{-1}}\left( \frac{{{X}_{C}}}{R} \right)$

The above result shows that in an AC circuit containing capacitor and resistor in series, voltage lags behind the current by phase angle$\phi $.

**AC circuit containing inductor, capacitor and resistor in series. (LCR in series).**

Let us consider an inductor of inductance L, a capacitor of capacitance C and a resistor of resistance R are connected in series with an AC source of emf, E = E_{o} sin$\omega t$. Let V_{L}, V_{C} and V_{R} be the potential across inductor, capacitor and resistor, respectively.

In AC circuit containing resistor only, the current and voltage will be in the same phase. The voltage V_{R} is represented by OA in the figure.

In AC circuit containing inductor only, the voltage leads the current by phase angle $\frac{\pi }{2}$ , the voltage V_{L} is represented by OY’ in the figure and in AC circuit containing capacitor only, the voltage lags behind the current by phase angle $\frac{\pi }{2}$, the voltage V_{C} is represented by OY in the figure, The voltage V_{L} and V_{C} are in opposite direction. If (V_{L} > V_{C}) then the resultant voltage of V_{L} and V_{C} i.e., (V_{L} – V_{C}) is represented by OC.

Let us complete a parallelogram OABC. Now the diagonal OB represents the effective voltage E of the LCR series circuit.

From $\Delta $OBA,

(OB)^{2} = (OA)^{2} + (AB)^{2}

or, (OB)^{2} = (OA)^{2} + (OC)^{2}

or, E^{2} = V_{R}^{2} + (V_{L} – V_{C})^{2}

or, E^{2} = I^{2}R^{2} + (I X_{L} – I X_{C})^{2}

or, $\frac{{{E}^{2}}}{{{I}^{2}}}$ = R^{2} + (X_{L} – X_{C})^{2}

or, Z^{2} = R^{2} + (X_{L} – X_{C})^{2} (⸪ Z = $\frac{E}{I}$ is impedance of AC circuit i.e. total resistance offered by AC circuit)

$\therefore $ Z = $\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

**Calculation of phase angle:**

From $\Delta $OAB,

Tan$\phi $ = $\frac{AB}{OA}$ = $\frac{OC}{OA}$

or, Tan$\phi $ = $\frac{{{V}_{L}}-{{V}_{C}}}{{{V}_{R}}}$

or, Tan$\phi $ = $\frac{I{{X}_{L}}-I{{X}_{C}}}{IR}$

or, Tan$\phi $ = $\frac{{{X}_{L}}-{{X}_{C}}}{R}$

$\therefore \phi ={{\tan }^{-1}}\left( \frac{{{X}_{L}}-{{X}_{C}}}{R} \right)$

**Special cases:**

I. If X_{L} > X_{C}

In this case, the value of phase angle$\phi $ is positive i.e. voltage leads the current by phase angle$\phi $.

II. If X_{L} = X_{C}

In this case, the value of phase angle$\phi $ is O. i.e. voltage and current will be in the same phase.

III. If X_{C} > X_{L}

In this case, the phase angle$\phi $ is negative i.e. voltage lags behind the current by phase angle$\phi $.

**Power consumed by AC circuit (LCR in series):**

*Fig: AC circuit containing inductor, capacitor and resistor in series*

Let an inductor of inductance L, a capacitor of capacitance C and a resistor of resistance R are connected in series with an AC source. Let I and E be the rms value of current and emf, respectively. Let us consider the case in which current lags behind the voltage by phase angle$\phi $. The instantaneous power developed is

P_{ins} = I.E.

or, P_{ins} = I_{o} sin ($\omega t$ –$\phi $) . E_{o} sin$\omega t$

or, P_{ins} = I_{o} E_{o} (sin$\omega t$. cos$\phi $ – cos$\omega t$ sin$\phi $) sin$\omega t$

or, P_{ins} = I_{o} E_{o} (sin^{2$\omega t$}. cos$\phi $ – sin$\omega t$ cos$\omega t$ sin$\phi $)

Let dw be the small amount of work done in small time dt

i.e. dw = P_{ins} . dt ………………… (i)

Now, the total amount of work done in a complete cycle can be obtained by integrating equation (i) from 0 to T.

or, $\int\limits_{0}^{T}{{}}$ dw = $\int\limits_{0}^{T}{{}}$ P_{ins} . dt

or, W = I. E. $\int\limits_{0}^{T}{{}}$ (sin^{2$\omega t$} cos$\phi $ – sin$\omega t$ cos$\omega t$ . sin$\phi $) dt

or, W = I. E.$\left[ \cos \phi \int_{0}^{T}{\left( \frac{1-\cos 2\omega t}{2} \right)dt-\frac{\sin \phi }{2}\int_{0}^{T}{\sin 2\omega t.dt}} \right]$

or, W =$\frac{{{I}_{o}}E{}_{o}}{2}$$\left[ \cos \phi \left( \int_{0}^{T}{dt-}\int_{0}^{T}{\cos 2\omega tdt} \right)-\sin \phi .0 \right]$ $\left( \because \text{ }\int\limits_{0}^{T}{\sin 2\omega t\text{ dt =0}} \right)$

or, W =$\frac{{{I}_{o}}E{}_{o}}{2}$ [cos$\phi $ {T – O}] $\left( \because \text{ }\int\limits_{0}^{T}{\cos 2\omega t\text{ dt =0}} \right)$

or, W =$\frac{{{I}_{o}}E{}_{o}}{2}$ . cos$\phi $ . T

or, W = $\frac{{{I}_{o}}}{\sqrt{2}}$ . $\frac{{{E}_{o}}}{\sqrt{2}}$ cos$\phi $ T

or, W = I_{rms} . E_{rms} cos$\phi $ T

Now, the average power developed in complete cycle is

P_{avg} = $\frac{W}{T}$

$\therefore $ P_{avg} = I_{rms} E_{rms} . cos$\phi $

Where, cos$\phi $ = $\frac{R}{Z}$ is known as power factor.

**Special cases**

I. Resistor only** **

We have, P_{avg.} = I_{rms} E_{rms} cos$\phi $

or, P_{avg.} = I_{rms} E_{rms} cos O ($\phi $ = 0°)

or, P_{avg.} = I_{rms} E_{rms}

II. Inductor only

We have, P_{avg.} = I_{rms} E_{rms} cos$\phi $

or, P_{avg.} = I_{rms} E_{rms} cos $\left( \frac{-\pi }{2} \right)$ $\left( \because \phi =\frac{-\pi }{2} \right)$

$\therefore $ P_{avg.} = 0

III. Capacitor only

We have, P_{avg.} = I_{rms} . E_{rms} cos$\phi $

or, P_{avg.} = I_{rms} . E_{rms} cos$\left( \frac{\pi }{2} \right)$ $\left( \because \phi =\frac{\pi }{2} \right)$

or, P_{avg.} = 0

IV. L and R in series

We have, P_{avg.} = I_{rms} E_{rm}_{s} cos$\phi $

cos$\phi $ = $\frac{R}{Z}$

Z = $\sqrt{{{R}^{2}}+{{X}_{L}}^{2}}$

V. R and C in series

We have, P_{avg.} = I_{rms} E_{r}_{ms} cos$\phi $

cos$\phi $ = $\frac{R}{Z}$

Z =$\sqrt{{{R}^{2}}+{{X}_{C}}^{2}}$

**Electrical resonance:**

In an AC circuit containing L, C and R in series if X_{L} = X_{C}, the value of impedance becomes minimum and current becomes maximum. This condition is known as electrical resonance. In electrical resonance, current and voltage will be in same the phase.

For electrical resonance,

X_{L} = X_{C}

or, $\omega L$ = $\frac{1}{\omega C}$

or, 2$\pi $f_{o} L = $\frac{1}{2\pi {{f}_{o}}C}$

or, f_{o}^{2} = $\frac{1}{4{{\pi }^{2}}LC}$

or, f_{o} = $\frac{1}{2\pi }$ $\sqrt{\frac{1}{LC}}$

Where f_{o} is known as resonance frequency.

**Quality factor:**

Quality factor is defined as the ratio of potential difference across inductor or capacitor to the potential difference across resistor in resonance condition.

Q. factor = $\frac{P.d.\text{ }across\text{ }inductor}{P.d.\text{ }across\text{ }resistor}$

= $\frac{{{V}_{L}}}{V{}_{R}}$

= $\frac{I{{X}_{L}}}{IR}$

= $\frac{\omega L}{R}$

= $\frac{2\pi {{f}_{o}}L}{R}$

= $2\pi .\frac{1}{2\pi }\sqrt{\frac{1}{LC}}\frac{L}{R}$

Q. factor = $\frac{1}{R}$ $\sqrt{\frac{L}{C}}$

**OR**

Q. factor = $\frac{P.d.\text{ }acrossc\text{ }apacitor}{P.d.\text{ }across\text{ }resistor}$

= $\frac{{{V}_{C}}}{{{V}_{R}}}$

= $\frac{I{{X}_{C}}}{IR}$

= $\frac{1/\omega C}{R}$

= $\frac{1}{2\pi {{f}_{o}}CR}$

= $\frac{1}{2\pi .\frac{1}{2\pi }\sqrt{\frac{1}{LC}}.CR}$

Q. factor = $\frac{1}{R}$ $\sqrt{\frac{L}{C}}$

**Choke coil:**

A coil having a large number of turns and negligible internal resistance is known as a choke coil. It is a pure inductor.

Average power consumed in AC circuit containing inductor only is

P_{avg.} = I_{rms} E_{rms} cos$\phi $

$\because $ For inductor only,$\phi $ = $\frac{\pi }{2}$

P_{Avg.} = I_{rms} E_{rms} cos $\frac{\pi }{2}$

$\therefore $ P_{Avg.} = O

**Wattless current:**

In an AC circuit containing an inductor only or capacitor only, the value of phase angle is $\frac{\pi }{2}$ and average power consumed is O. Current in this case is known as wattles current.

**Wattful current:**

In an AC circuit containing resistor only, the average power consumed is I_{rms} E_{rms}. The current in this case is known as watt full current.

**Admittance:**

Admittance is defined as the reciprocal of impedance. It is denoted by Y.

Mathematically,

Admittance, Y = $\frac{1}{Impedance\text{ }(Z)}$

Y = $\frac{1}{\sqrt{R2+{{({{X}_{L}}-{{X}_{C}})}^{2}}}}$

Its unit is $\Omega $^{–1}.

**Alternating Current** **Formulae for Numerical Problems:**

1. The instantaneous value of alternating current is

I = I_{o} sin$\omega t$.

Where I_{o} is maximum (peak) value of current.

And the instantaneous value of alternating voltage is

E = E_{o} sin$\omega t$.

Where E_{o} is maximum (peak) value of voltage.

2. For certain phase angle ($\phi $)

E = E_{o} sin$\omega t$ and I = I_{o} sin ($\omega t\pm \phi $) where φ is phase angle between emf and current.

3.** **I_{rms} = $\frac{{{I}_{o}}}{\sqrt{2}}$ &** **E_{rms} = $\frac{{{E}_{o}}}{\sqrt{2}}$

4. Phase angle ($\phi $) between current and voltage in AC circuit containing

R | L | C | R & L | R & C | L,C,R |

$\phi $ = 0 | $\phi $ = – $\frac{\pi }{2}$π | $\phi $ = $\frac{\pi }{2}$ | Tan$\phi $ = $\frac{{{X}_{L}}}{R}$ | Tan$\phi $ =$\frac{{{X}_{C}}}{R}$ | Tan$\phi $ =$\frac{{{X}_{L}}-{{X}_{C}}}{R}$ |

i.e. I & E are in same phase | i.e. I lags behind the E by $\frac{\pi }{2}$ (i.e. by 90^{o}) | i.e. I leads the E by $\frac{\pi }{2}$(i.e. by 90^{o}) | i.e. I lags behind the E by $\frac{\pi }{2}$(i.e. by 90^{o}) | i.e. I leads the E by phase angle $\phi $. | For X_{L} > X_{C} i.e. I lags behind the E by phase angle$\phi $. |

**5.**** **Inductive reactance (i.e. resistance offered by inductor),

X_{L} =$\omega L$ = 2$\pi $f L

Capacitive reactance (i.e. resistance offered by inductor),

X_{c} = $\frac{1}{\omega c}$ = $\frac{1}{2\pi fC}$

6. Impedance of AC circuit i.e. total resistance offered by AC circuit

Z =$\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

7. Voltage drop across R, V_{R} = IR

Voltage drop across L, V_{L} = I X_{L}

Voltage drop across C, V_{C} = I X_{c}

8. Circuit Formula E = IZ (in AC) [E=V=I R_{eq} , for r = 0 (in D.C)]

9. Electrical Resonance

At resonance, I is max^{m} and hence Z is min^{m} as E= IZ

We have, Impedance, Z =$\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

i.e. at resonance, X_{L} = X_{c}

Resonating frequency, f_{o} = $\frac{1}{2\pi }$ $\sqrt{\frac{1}{LC}}$

10. Quality factor = $\frac{{{V}_{L}}\text{ }or\text{ }{{V}_{C}}}{{{V}_{R}}}$

Q. factor = $\frac{1}{R}$ $\sqrt{\frac{L}{C}}$

11. Average power consumed in AC circuit

P_{avg} = I_{rms} E_{rms} . cos$\phi $

12. Power factor, cos$\phi $ = $\frac{R}{Z}$ = $\frac{R}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}}$

13. A.C. Circuit containing inductor with some internal resistance (${{r}_{L}}$)

(i) Impedance of AC circuit, Z = $\sqrt{{{(R+{{r}_{L}})}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

(ii) Impedance of AC circuit containing L & R, Z =$\sqrt{{{(R+{{r}_{L}})}^{2}}+{{X}_{L}}^{2}}$

(iii) Total resistance offered by inductor, Z_{L} = $\sqrt{{{r}_{L}}^{2}+{{X}_{L}}^{2}}$

(iv) Voltage drop across inductor, V_{L}= I Z_{L}

(v) Phase angle ($\phi $), Tan$\phi $ = $\frac{{{Z}_{L}}}{R}$

Also Read: DC Circuit Class 12 Notes

Biraj RegmiLovely sir! You are the best

Rambabu MandalTq so much….l can learn about ac.