AL Alternating Current P.160. P.161 P.163 2πf = 50 π f = 25 X -1

AL Alternating Current

P.160

P.160

time

I

1 cycle

P.163

2πf = 50 π

f = 25

X

-1 <= sin () <= 1

-2√2 <= I <= 2√2

Irms = Io / √2

Irms = 2√2 / √2

Irms = 2

P.163

In one cycle, power dissipated is

P = (5)2(1x10-3)R + (-3)2(1x10-3)R

P = 34 R(1x10-3)

Pdc = = I2 R(2x10-3)

34 R(1x10-3) = I2 R(2x10-3)

I = 4.1 (A)

P.163

Irms = 1 / √2

Energy dissipated in sine wave is Irms2 R T = (1/2) R T

Energy dissipated in square wave is Io2 R (T/2) + 0 = (1/2) Io

2 R T

Io= 1 (A)

P.163

Same heating effect => using rms

For whole sine wave, Irms = 2 / √2 => Irms2 R= 2 R

For rectified case, the ohmic loss will be halved. P= R = Irms(new)2 R

Irms(new) =1

P.175

Vrms= Irms √(R2 + Xc2)

When f ↑, (1/(2πf C)) ↓, Z ↓, Irms ↑

Irms= Vrms (√(R2 + (1/(2πf C))2))-1

Vc= Irms Xc

Vc= Vrms ( Xc / √(R2 + Xc2))

Vc= Vrms ( 1 / √(1 + (2πf C R)2))

When f ↑, Vc ↓

P.175

L dI/dt = N dΦ/dt

As VL = -L dI/dt and VR = √(Vo2 – VL

2)

X

X

L I = N Φ = N B A = N A (μ N I / l)

L = N2 A μ / l

When insert iron core, μ↑, L ↑

When L ↑, VL↑, VR ↓

As XL = 2 πf L

When f ↑, VL↑, VR ↓

P.175

VL – VC = 4 – 1 = 3

tan φ = (VL – VC ) / VR = 3/4

VR = 4

φ = 37o

P.176

In RC circuit, Ic leads Vc by π/2

Y1 measures VR , and VR is in phase with Ic, so VR leads Vc by π/2

Y2 measures VC but reverse in direction

P.176

In RL circuit, Z=√(R2 + (ωL)2)

After adding C, the circuit will be LRC circuit

Peak value of bulb Vb=V(R/Z) =V(R/√(R2 + (ωL)2))

When f ↑, Vb↓, Brightness↓as resistance is constant

X

When inserting iron core, L ↑, Vb↓, Brightness↓as resistance is constant

X

With suitable C, resonance could be occurred, so the total impedance is minimized and the current will be maximum while VR is as the same as the p.d. of source

P.181

From the figure, current should be flowing towards the lower plate of capacitor

The current flows in clockwise direction. There are 2 possible cases.

1) The capacitor is charging. The upper plate is negative while the lower plate is positive.

2) The capacitor is discharging. The upper plate is positive while the lower plate is negative.

P.181

Electric field CANNOT form closed loops.

P.189

+

-

a.c.input R

+

-

d.c.output

+

-a.c.input R

+

-

Nooutput

P.189

+

-a.c. input

R +

-

d.c. output

+

-

+

-a.c. input

R +-

d.c. output+

-

P.189

a.c.input

+

--

R

+

d.c.output

a.c.input

+

-

-

R

+

d.c.output

P.190

+

-

Rectifiedunsmoothed

p.d.R

+

-

d.c.output

C

P.191

+

-

Rectifiedunsmoothed

p.d.R

+

-

d.c.output

C 2C 1

L

+

-

Rectifiedunsmoothed

p.d.

Steadyd.c.

output

C 2

C 1

L

V 1

Ripplep.d.

P.191

-

+

mA

P.192

-

+

d.c.

+

-

1kW

1W50 V2000 m F

a.c.

1 A

12 V5 A

5 A

12 V0.3 A

P.192

Irms = Io / √2 Idc = Io (2 / π)

Idc = 1 = Io (2 / π)

Io = π / 2 = 1.57

Irms = 1.57 / √2

Irms = 1.11

P.192

This is RC circuit.

Xc = 1 / (ωC) = 1/(2πf C)

R = V/I

Z = √(R2 +Xc2)

Vs = I Z = I √(R2 +Xc2)

Vs = I √((V/I)2 + 1/(2πf C)2)

Vs = (0.3) √((2.5/0.3)2 + 1/(2π(50) (1000x10-6))2)

Vs =

Vc = I Xc = I /(2πf C)

Vc = (0.3) /(2π(50) (1000x10-6))

Vc =

VR = I R = 2.5 (V)

P.192

This is LR circuit.

XL = ωL = 2πf L

R = 80+420 = 500

Z = √(R2 +XL2)

Vs = I Z = I √(R2 +XL2)

Vs = I √(R2 + (2πf L)2)

240 = I √(5002 + (2π(50) (2))2)

I = 0.3 (A)

tanφ =XL / R

tanφ = ωL /R

φ = tan-1 (2π(50) (2) / 500

φ = 51.5o

P.193

At resonance, Xc = XL

1/(ωC) = ωL

C = 1/(ω2 L)

C = 1/((2πf)2 L)

C = 0.12665n (F)

Vc = I Xc = (V/R) Xc

Vc = (V/R)(1/(2πf C))

Vc = VL = 12.566 (V)

Q = Xc / R =(1/(2πf C))/R

Q = 125.66

P.193

a) P = I2 R

R = (0.4 x 103) / (2)2 = 0.1 kΩ

b) Vs = I √(R2 +XL2)

240 = (2) √(1002 +(2π(50) L)2)

L =0.211 (H)

c) VL leads both IL, VR

tan φ = XL/Rlamp

In RC circuit, VC lags both IC, VR

tan φ’ = XC/R50

To compensate the phase lead,tan φ’ = tan φ =>XC/R50= XL/Rlamp

(1/ωC)/50= ω (0.211)/100

C= 96μF

P.193

In RC circuit, I leads both Vs and Vc.

Z = √(R2 +1/(ωC)2)

As ω increases, Z increases

Power factor = cos φ= > increases

As ω increases, Xc decreases => less important => φdecreases

P.193

Inductor will not consume any power

All power dissipated comes from the ohmic loss

Power loss P = I2 R = (2)2 5 = 20 (W)

P.193

Z2 = R2 + X2

All power dissipated comes from the ohmic loss

Power loss P = I2 R = (2)2 5 = 20 (W)

132 = R2 + 122

R = 5

P.194

A) Energy consumed by R only or stray C

B) Energy is released from L, while energy is absorbed by C and vice versa

C) By equation, it contains frequency f.

D) In series circuit, all current are the same.

AL Alternating Current P.160. P.161 P.163 2πf = 50 π f = 25 X -1

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Transcript of AL Alternating Current P.160. P.161 P.163 2πf = 50 π f = 25 X -1