Circuits With Resistance And Inductance L-3 Circuits with Resistance and Inductance
→ \rightarrow → → \rightarrow → Circuits with Resistance and Inductance → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Faraday's Law of Electromagnetic Induction
Faraday's Law
ε = − N d Q d t \varepsilon=-N\frac{dQ}{dt} ε = − N d t d Q
N = Number of turns N = \text{Number of turns} N = Number of turns
Φ = Flux through each turn \Phi = \text{Flux through each turn} Φ = Flux through each turn
ε = ∫ B ⃗ . d s ⃗ \varepsilon = \int\vec{B}.\vec{ds} ε = ∫ B . d s Lenz's Law .
→ \rightarrow → → \rightarrow → Faraday's Law of Electromagnetic Induction → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Lenz's Law
The direction of current is always such that the magnetic field produced by it, opposes the change that produced it.
Φ B = B A cos θ ( t ) \Phi_B= \text{B A}\cos\theta (t) Φ B = B A cos θ ( t )
= B A cos ( ω t ) \text{B A} \cos (\omega t) B A cos ( ω t )
ε = − N d ϕ B d t = \varepsilon=-N \frac{d \phi_B}{d t} = ε = − N d t d ϕ B = ω sin ω t ~\omega \sin \omega t ω sin ω t
= ϵ 0 sin ω t . =\epsilon_0 \sin \omega t . = ϵ 0 sin ω t .
Circuits with Resistance and Inductance → \rightarrow → → \rightarrow → Lenz's Law → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Alternating Voltage
Potential V = V m sin ω t V = V_m \sin {\omega t} V = V m sin ω t
T : Period
V(t + T ) = V ( t ) \text{V(t + T )} = V (t) V(t + T ) = V ( t )
Linear Frequency, f = 1 T f = \frac{1}{T} f = T 1
ω = 2 π f = 2 π T . \omega = 2\pi f = \frac{2\pi}{T}. ω = 2 π f = T 2 π .
Faraday's Law of Electromagnetic Induction → \rightarrow → → \rightarrow → Alternating Voltage → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Alternating Voltage
V = V m sin ω t V = V_m \sin\omega t V = V m sin ω t
I = ( V m R ) sin ω t I=\left(\frac{V_m}{R}\right) \sin \omega t I = ( R V m ) sin ω t
= I m sin ω t . = I_m \sin \omega t. = I m sin ω t .
(I m = I_m = I m =
Lenz's Law → \rightarrow → → \rightarrow → Alternating Voltage → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Alternating Voltage
When the voltage becomes maximum the current becomes maximum.
Alternating Voltage → \rightarrow → → \rightarrow → Alternating Voltage → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Phasor Diagram
t = 0 (Reference Line)
∣ O A → ∣ = V m |\overrightarrow{O A}|=V_m ∣ O A ∣ = V m
∣ O C → ∣ = I m |\overrightarrow{O C}|=I_m ∣ OC ∣ = I m
V ( t ) = V m sin ω t V (t) = V_m \sin \omega t V ( t ) = V m sin ω t
I ( t ) = I m sin ω t I (t) = I_m \sin \omega t I ( t ) = I m sin ω t
For a purely resistive circuit, the current is in Phase with the voltage.
Alternating Voltage → \rightarrow → → \rightarrow → Phasor Diagram → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Average Force
The average of F (t) over a cycle,
⟨ F ( t ) ⟩ = 1 T ∫ 0 T F ( t ) d t \langle F(t) \rangle = \frac{1}{T} \int_{0}^{T} F(t) dt ⟨ F ( t )⟩ = T 1 ∫ 0 T F ( t ) d t
I ( t ) = I m . sin ( ω t ) I (t) = {I_m}. \sin (\omega t) I ( t ) = I m . sin ( ω t )
⟨ I ( t ) ⟩ = I m . 1 T ∫ 0 T sin ω t d t \langle I(t) \rangle = {I_m}.\frac{1}{T} \int_{0}^{T} \sin \omega t dt ⟨ I ( t )⟩ = I m . T 1 ∫ 0 T sin ω t d t
= I m 1 ω T [ 1 − cos ω T ] {I_m} \frac{1}{\omega T} [1 - \cos \omega T] I m ω T 1 [ 1 − cos ω T ]
ω T = 2 π = 0 \omega~T = 2\pi = 0 ω T = 2 π = 0
Alternating Voltage → \rightarrow → → \rightarrow → Average Force → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Average Force
Also valid for functions like sin 2 ω t . \sin 2 \omega t. sin 2 ω t .
cos ω t , cos 2 ω t \cos \omega t, \cos 2 \omega t cos ω t , cos 2 ω t
⟨ sin 2 ω t ⟩ = ? \langle \sin^2 \omega t \rangle = ? ⟨ sin 2 ω t ⟩ = ?
⟨ sin 2 ω t ⟩ = 1 T ∫ 0 T sin 2 ω t d t \langle \sin^2 \omega t \rangle =\frac{1}{T}\int_{0}^{T} \sin^2 \omega t~dt ⟨ sin 2 ω t ⟩ = T 1 ∫ 0 T sin 2 ω t d t
= 1 T ∫ 0 T 1 − cos 2 ω t 2 d t = 1 2 \frac{1}{T} \int_{0}^{T} \frac{1-\cos 2\omega t}{2} dt = \frac{1}{2} T 1 ∫ 0 T 2 1 − c o s 2 ω t d t = 2 1
( ∵ ⟨ cos 2 ω t ⟩ = 0 ) (\because \langle\cos 2 \omega t \rangle =0) ( ∵ ⟨ cos 2 ω t ⟩ = 0 )
Phasor Diagram → \rightarrow → → \rightarrow → Average Force → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Root Mean Square Current
Average Current is Zero
P a v = ⟨ I 2 R ⟩ P_{av} = \langle I^2 R \rangle P a v = ⟨ I 2 R ⟩
= I m 2 R ⟨ sin 2 ω t ⟩ I_ m^2 R \langle \sin^2 \omega t \rangle I m 2 R ⟨ sin 2 ω t ⟩ I m 2 R 2 \frac{I_m^2 R}{2} 2 I m 2 R
Root mean square current
I r m s = ⟨ I 2 ( t ) ⟩ = I m 2 . I_ {rms}=\sqrt{\langle I^2(t) \rangle} = \frac{I_m}{\sqrt{2}}. I r m s = ⟨ I 2 ( t )⟩ = 2 I m .
V r m s = V m 2 {V_ {rms}} = \frac{V_m}{\sqrt{2}} V r m s = 2 V m
P a v = I r m s 2 R {P _ {av}} = I^2_ {rms}~R P a v = I r m s 2 R
Average Force → \rightarrow → → \rightarrow → Root Mean Square Current → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 House Hold AC
House hold AC supply in India
220 − 240 V 220 - 240 V 220 − 240 V
v = 50 H z . v = 50 Hz. v = 50 Hz .
USA 120 v , 60 H z 120 v, 60 Hz 120 v , 60 Hz
Average Force → \rightarrow → → \rightarrow → House Hold AC → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Current Amplitude
Alternating Source applied to a purely inductive load No Resistance.
Back emf = − L d I d t -L\frac{dI}{dt} − L d t d I
V(t) − L d I d t = 0 -L\frac{dI}{dt}=0 − L d t d I = 0
d I d t = V ( t ) L = ( V m L ) sin ω t \frac{dI}{dt}=\frac{V(t)}{L}=(\frac{V_m}{L})\sin \omega t d t d I = L V ( t ) = ( L V m ) sin ω t
I (t)= ∫ V m L sin ω t = − V m L ω cos ω t \int\frac{V_m}{L} \sin \omega t = -\frac{V_m}{L\omega} \cos \omega t ∫ L V m sin ω t = − L ω V m cos ω t
=V m L ω sin ( ω t − π 2 ) \frac{V_m}{L\omega} \sin(\omega t-\frac{\pi}{2}) L ω V m sin ( ω t − 2 π )
V m L ω \frac{V_m}{L\omega} L ω V m
( ω t − π 2 ) (\omega t-\frac{\pi}{2}) ( ω t − 2 π ) = Phase lag of π / 2 \pi/2 π /2
Root Mean Square Current → \rightarrow → → \rightarrow → Current Amplitude → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Inductive Reactance
I(t)=V m ( ω L ) sin ( ω t − π 2 ) \frac{V_m}{(\omega L)} \sin (\omega t-\frac{\pi}{2}) ( ω L ) V m sin ( ω t − 2 π )
( ω L ) (\omega L) ( ω L )
X L = ω L X_L = \omega L X L = ω L
House Hold AC → \rightarrow → → \rightarrow → Inductive Reactance → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Inductive Circuit
For inductive circuit, current Lags the voltage (in phase by π 2 ) \frac{\pi}{2}) 2 π )
Current Amplitude → \rightarrow → → \rightarrow → Inductive Circuit → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Phasor Diagram (Inductive Circuit)
Inductive Reactance → \rightarrow → → \rightarrow → Phasor Diagram (Inductive Circuit) → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Instantaneous Power
P = I V P = IV P = I V
= I m sin ( ω t − π 2 ) V m sin ( ω t ) I_m \sin (\omega t - \frac{\pi}{2}) V_m \sin (\omega t) I m sin ( ω t − 2 π ) V m sin ( ω t )
= − I m V m . cos ω t sin ω t -I_m ~ V_m. \cos \omega t \sin \omega t − I m V m . cos ω t sin ω t
= − I m V m 2 sin 2 ω t - \frac{I_m V_m}{2} \sin 2 \omega t − 2 I m V m sin 2 ω t
⟨ P ⟩ \langle P \rangle ⟨ P ⟩
Inductive Circuit → \rightarrow → → \rightarrow → Instantaneous Power → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Inductive Circuit
For an inductive circuit
⟨ P ⟩ \langle P \rangle ⟨ P ⟩
Compare with a Purely resistive circuit
⟨ P ⟩ = I r m s 2 R . \langle P \rangle = I_{rms}^2 R. ⟨ P ⟩ = I r m s 2 R .
Phasor Diagram (Inductive Circuit) → \rightarrow → → \rightarrow → Inductive Circuit → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Phasor Diagram
V (t) in Volts
I (t) in ampere
From T / 4 T / 4 T /4 T / 2 I > 0 , d I d t > 0 V ( t ) > 0 T / 2 \quad I>0, \frac{d I}{d t}>0 \quad V(t)>0 T /2 I > 0 , d t d I > 0 V ( t ) > 0
P = I V > 0 P = I V > 0 P = I V > 0
From T / 2 to 3 T / 4 I > 0 T / 2\text{ to}\quad 3 T / 4 \quad I>0 T /2 to 3 T /4 I > 0 d I d t < 0 ; V ( t ) < 0 \frac{d I}{d t}<0 ; V(t)<0 d t d I < 0 ; V ( t ) < 0
P = I V < 0 P = I V < 0 P = I V < 0
From 3 T 4 \frac{3 T}{4} 4 3 T T I < 0 , d I d t < 0 : V ( t ) < 0 T \quad I<0, \frac{d I}{d t}<0: V(t)<0 T I < 0 , d t d I < 0 : V ( t ) < 0
P > 0 P > 0 P > 0
Instantaneous Power → \rightarrow → → \rightarrow → Phasor Diagram → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Example
48 m H inductor connected to 240V, 50 Hz supply. What is I r m s I_{rms} I r m s
X L = ω L = 2 π × 50 × 48 × 10 − 3 X_L = \omega L = 2\pi \times 50 \times 48 \times 10^{-3} X L = ω L = 2 π × 50 × 48 × 1 0 − 3
= 4.8 π = 15.08 Ω = 4.8 \pi = 15.08 \Omega = 4.8 π = 15.08Ω
I r m s = 240 16 = 16 A . I_{rms} = \frac{240}{16}= 16 A. I r m s = 16 240 = 16 A .
Inductive Circuit → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Circuits With Resistance And Inductance L-3 Thank You
Phasor Diagram → \rightarrow → → \rightarrow → Thank You → \rightarrow → → \rightarrow →
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Circuits With Resistance And Inductance L-3 Circuits with Resistance and Inductance $\rightarrow$ $\rightarrow$ Circuits with Resistance and Inductance $\rightarrow$ Faraday's Law of Electromagnetic Induction $\rightarrow$ Lenz's Law
