Alternating voltage in primary

Alternating emf in secondary

Soft Iron

Increase magnetic field

Help flux linkage

$\phi$ = Flux through each turn of primary.

$N_p$ = no of turns in primary

$V_p = -N_p$ . $\frac {d\phi}{dt}$

$V_s = -N_s$ . $\frac {d\phi}{dt}$

$\frac {v_s}{v_p}= \frac{N_s}{N_p}$

Ns = No of turns in secondary

No flux leakage

$N_s>N_p$ : step up transform

$N_s<N_p$ : step Down

All power is transformed

$I_p V_p = I_s V_s$

$I_p = I_s. \frac {V_s}{V_p}$

= $I_s.\frac {N_s}{N_p}$

Nail R = .004 $\Omega$

$I = \frac{240}{.004}$ = 60,000A

Q = $I_s^2 R \times t$

$I_s$ is 500A

$V_s$ = $I_s \times R_s$ = 500 $\times$ .004 = 2V

$\frac {N_p}{N_s}$ = $\frac {120}{1}$

$I_p . N_p = N_s . I_s$

$I_p = \frac {1}{120} \times 500$ = 4.16 A

Power delivered

$I_s^2 R = (500)^2 \times .oo4$ = 1000w

In practice, all power is not transforred.

1. Flux linkage is not complete $\rightarrow$ self reactance

(redyce it by tight coupling)

2. Resistance of windings - not zero

$R_p, X_p$ : Resistance & reactance of primary

$R_s, X_s$ : Secondary

$Z_p = \sqrt {R_p^2 + X_p^2} : Z_s = \sqrt {R_s^2 + X_s^2}$

Induced emf across primary is not $V_p$ byut reduced by $I_p Z_p$

Secondary: not $V_s$ but reduced by $I_s Z_s$

Power transfer : Not complete

a) Eddy losses is the iron core

b) Repeated magnetization - Hystersis loss

Reduce eddy current losses

we use laminated core

Eddy current for a single block

effect of lamination

Reduc eddy current losses

we use laminated core

Hysteresis loss: use of soft magnetic material with law hystersis

[ Si- steel, steel alloys , Mn-Bn ferrites]

Arises due to resistance of the windings.

Loss in primary $I_p^2 R_p$

Loss in secondary $I_s^2 R_s$

Losses depends on load.

Reduction is possible

Use of thick wires as winding wires.

$N_p$ = 200 $N_s$ = 10

$V_p$ = 240V Load $R_s = 20 \Omega$

step down by a factor of 20

$V_s = \frac {240}{20}$= 12V

$I_s = \frac {12}{20}$= 0.6V

$I_s V_s = I_p V_p$

0.6 $\times 12 = I_p \times 240$

$I_p$ = 0.03A

$P_{lost} = I^2 R_c$

$R_c$ = cable resistance

Reduce $p_{loss} \rightarrow$ I should be as small as possible

P = IV : V should be large

$P_{out} = 1MW = 10^6W$ =IV

$P_{lost} = T^2 R_c$

$\frac {P_{lost}}{P{out}} = \frac P{out}{V^2}.R_V$

Suppose $R_c = 10\Omega$, $P_{out} = 10^6W$, V = 20 KV

$\frac {P_{lost}}{P_{out}} = \frac {10^6}{(2 \times 10^4)^2} \times 10 = 0.025$

2.5 % loss, if v = 200 kv : 0.025 %

Rotating cell in B.

$V = V_m cos \omega t$

T = $\frac {1}{f}$ = $\frac {2 \pi}{w}$

For a purely resistive circuit

current is in phase with the voltage

$V(t) = V_m sin \omega t$

$I(t) = I_m sin \omega t$

$I_m = \frac {V_m}{R} : I_{rms} = \frac {I_m}{sqrt 2}$

P(t) = I(t) V(t)

$< P(t) > = I_{rms}. R = \frac {V^2_{rms}}{R}$

$V(t) = V_m sin \omega t$

$Q(t) = C V_m sin \omega t$

$I(t) = I_m sin (\omega t + \frac {\pi}{2})$

Current leads the voltage by $\frac {\pi}{2}$

$I_m = \frac {V_m}{X_c}$

$X_c$ = capacitive rectance

$= \frac {1}{\omega c} . \Omega$

$V = V_m sin \omega t$

I = $\frac {V_m}{L_\omega} .sin (\omega - \frac {\pi}{2})$

current lags the voltage by $\frac {\pi}{2}$

$I_m = \frac {V_m}{L_\omega}: L_\omega = X_L$ inductive reactance

ELI the ICE man.

V = I Z

Z = Impedance

= $\sqrt {R^2 + (X_c - X_L)^2}$

= $\sqrt {R^2 + X^2}$

V = $V_m sin \omega t$

I = $I_m sin (\omega t + \varphi)$

P = IV

= $I_m sin(\omega t + \varphi) V_m sin \omega t$

= $I_m V_m sin^2 \omega t + cos \varphi + I_m V_m sin \omega t cos \omega t sin \varphi$

$<P> = \frac {I_m V_m}{2} . cos \varphi$ - power factor

Z = $\sqrt {R^2 + (X_L - X_c)^2}$

For $X_L = X_c$ : Z has a minimum

By varrying $\omega$, we get a maximum amplitude where

$\omega = \omega_0 = \frac {1}{\sqrt LC}$ Resonant frequency

$R_1 > R_2 > R_3$