ALTERNATING CURRENT CIRCUITS - Florida Atlantic...

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CHAPTER 29 ALTERNATING CURRENT CIRCUITS ac generators Resistance in ac circuits and rms values Introduction to phasors Inductors in ac circuits Capacitors in ac circuits R-C-L circuits with an ac generator Resonance and tuned circuits Transformers ! A ! B φ ε b b φ = ωt ω is the angular velocity The flux through coil is Φ = ! B ! A = BA cos φ = BA cos( ωt). dΦ dt = −ωBA sin ( ωt), so ε = dΦ dt = ωBA sin ( ωt). ε Φ t Time for one revolution (T) brushes slip rings

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Page 1: ALTERNATING CURRENT CIRCUITS - Florida Atlantic …courses.science.fau.edu/~jordanrg/phy2049/chapter_29/notes_29.pdf · ALTERNATING CURRENT CIRCUITS ... Alternating potential difference

CHAPTER 29

ALTERNATING CURRENT CIRCUITS

• ac generators

• Resistance in ac circuits and rms values

• Introduction to phasors

• Inductors in ac circuits

• Capacitors in ac circuits

• R-C-L circuits with an ac generator

• Resonance and tuned circuits

• Transformers

! A

! B φ

ε

b

b

φ = ωtω is the angular velocity

The flux through coil is

Φ =! B •! A = BAcosφ = BAcos(ωt).

∴dΦ

dt = −ωBA sin(ωt),

so ε = −dΦdt = ωBAsin(ωt).

ε

Φ

t

Time for one revolution (T)

brushes

sliprings

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Question 29.1: A coil with area 2.0 m2rotates in a 0.01 T

magnetic field at a frequency of 60 Hz. How many turns are needed to generate a peak voltage of 160 V?

For one turn we have ε = ωBA sin(ωt). So for N turns,

ε = NωBAsin(ωt).Peak voltage occurs when sin(ωt) = 1,

i.e., εpeak = NωBA.

Remember ω = 2πf , where f is the frequency.

∴N =

εpeakωBA

=

160 V(2π × 60 Hz)(0.01 T)(2.0 m2)

= 21 turns.

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~

Frequency f = ω2π

If w ⇒ rad/s then f ⇒ Hertz (Hz)With f = 60 Hz, ω = 376 rad/s

Periodic time T = 1f = 16.7ms

amplitude

instantaneousvalue

v(t) = Vcos(ωt)

v(t) = Vsin(ωt)

−V T

2 T

3T

2 2T

v( t) V

t

−V

v( t) V

t

Alternating potential difference ⇒ alternating current: i(t) = i = Icos(ωt)

With household a.c. (60Hz) the current reverses direction

every 1120 s, i.e., 120 times every second!

∴T = 1

f = 1(60 Hz) = 1.67 ×10−2s.

Distance travelled in one-half cycle is ~ vD × T2

= (10−4 m/s) × (0.83×10−2s) = 8.3×10−7 m.

(~ 3000 atom diameters)Not very far!!

T

2 T

3T

2 2T

i(t)

t

++++ t

t +T

2

− I

I

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The voltage and current are in-phase.

Resistor in an ac circuit

Instantaneous current i(t) = i = Icos(ωt)Instantaneous p.d. (voltage) across R is

vR = iR = IR cos(ωt) = VR cos(ωt),which is the same as the emf of the generator.

R

~i

vR

I = VR

R

VR = IR vR(t) & i(t)

Max vR

Max i

vR(t)

t i(t)

angularvelocity

A “Phasor” is a rotating vector. Phasors can be added just like ordinary vectors.

Voltage phasor: vR = IR cos(ωt) = VR cos(ωt)

! V R

vR

ωt

t

vR(t)

VR cos(ωt)

vR

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Current and voltage phasors:

i = Icos(ωt)

vR = IR cos(ωt) = VR cos(ωt)

i

! V R

! I

vR

ωt

t

vR(t) & i(t)

VR cos(ωt)

Icos(ωt)

vRi

Note: the current and voltage are in-phase.

Power dissipated in the resistorThe instantaneous power pR = vRi.

But i = Icos(ωt) and vR = IR cos(ωt)

∴pR = I2R cos2(ωt).

The instantanous power varies so we define the average power over one complete period:

P av =

Energy dissipated in one periodtime of one period

=

1T

I2

0

T∫ Rcos2(ωt)dt =

I2RT

cos2(ωt)dt0

T∫

=

I2RT

ωt2

+sin(2ωt)

4⎡ ⎣ ⎢

⎤ ⎦ ⎥ 0

T=

12

I2R

i.e., half the maximum power.

vR

pR

it

I2R

P av

T

2 T 0

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Let’s look for a moment at: i2 = I2 cos2(ωt)

The average “squared” current (over one period) is:

i2 =

Area under curve of i2(t) over one period Time of one period

=

1T

I2

0

T∫ cos2(ωt) dt = I2

2 .

We define the root mean square (rms) current as:

Irms = i2 = I22 = I

2 ( = 0.707I)

... and similarly: Vrms = V2 ( = 0.707V).

Average power: P av = I2R2 = Irms

2 R

= VrmsIrms =Vrms

2

R .

t 0

T

2 T

i2(t)

I2

hatched area = I2T

i2

Question 29.2: A current, i(t), varies with time as a square wave, as shown above, what is its rms value?

i(t)

I

−I

t

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i(t)

I

−I

t

Sketch i2(t) versus t and we get ...

By definition Irms = i2 , where the average square

value is

i2 =

Area under the i2(t) curve over one period Time of one period

=

I2TT

= I2.

∴Irms = i2 = I2 = I.

= I2T

0 T

2 3T

2 T 2T

i2(t)

I2 Question 29.3: A 100 W lightbulb is screwed into a standard 120 V(rms) socket. Find

(a) the rms current,(b) the peak current, and(c) the peak power.

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(a) Given : P av = 100 W and Vrms = 120 V.

But P av = VrmsIrms.

∴Irms =

P avVrms

=100 W120 V

= 0.833 A.

(b) Also Irms =

Imax2

and Vrms =

Vmax2

.

∴Imax = 2 × Irms = 1.18 A

and Vmax = 2 × Vrms = 169.7 V.

(c)

The peak (maximum) power Pmax = ImaxVmax

= (1.18 A)(169.7 V) = 200 W,i.e., maximum power = twice average power.

Note: the power pulses occur at 2ω .

~

100 W

120 V

Imax

Pmax Vmax

t

Inductor in an ac circuit

Instantaneous current: i = Icos(ωt)Instantaneous potential difference across inductor:

vL = L

didt

= −IωLsin(ωt),

which is the same as the emf of the generator.

But − sin(ωt) = cos ωt + π2( ) ... check it out!

∴vL = IωLcos ωt + π2( )

Reactance ( XL) is like “ac resistance” but it is not

constant; it depends on frequency.

∴vL = IXL cos ωt + π2( ).

Therefore, in terms of angle, vL is 90! ahead of the

current!

XL = ωL“reactance” (Ω) “phase angle”

~i

L

vL

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i = Icos(ωt)

vL = IXL cos(ωt + 90!) = VL cos(ωt + 90!),

where VL = IXL ( = IωL) i.e., the maximum voltage.

vL is 1

4 of a cycle ( 90!) ahead of i.

vL (t)

VL = IωL

90!

t

I

i(t)

Corresponding phasor diagram of the voltage and current for an inductor.

90! " V L

" I

ωt

i = I cos(ωt) vL

vL = IXL cos(ωt + 90!)

Progression of the voltage and current phasors for an inductor in series with an ac generator.

The voltage ‘leads’ the

current by 90!. The plots

show how the projections of the phasors on the x-axis, i.e., the instantaneous values of the current and voltage, vary with time.

90!

45!

135!

" V L

" I

vL i

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Power in an ideal inductor in an a.c. circuit:The instantaneous power pL = vLi.

But i = Icos(ωt) and vL = IωLcos(ωt + 90!)

∴pL = I2ωLcos(ωt)cos(ωt + 90!).

The average power (over one cycle):

P av =

1T

I2ωLcos(ωt)cos(ωt + 90!)0

T∫ dt = 0.

... You can tell that from the graph ...... the energy stored in the inductor in one quarter-cycle is released in the next quarter-cycle so the average power over one cycle is zero.

i

vL

I

VLt

pL

t0

T2 T

Capacitor in an ac circuit

Instantaneous current: i = Icos(ωt)

=dq

dt .

Instantaneous charge on the capacitor:

q = idt∫ = I cos(ωt)dt∫ =

Iωsin(ωt).

Therefore, the instantaneous potential difference across the capacitor is:

vC =

qC=

IωC

sin(ωt),

which is the same as the emf of the generator.

But sin(ωt) = cos ωt − π2( ).

∴vC =

IωC

cos ωt − π2( ).

i.e., vC = IXC cos ωt − π2( ).

Therefore, in terms of angle, vC is 90! behind the current.

~i

vC

C

XC =

1ωC

reactance (Ω) phase angle

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i = Icos(ωt)

vC = IXC cos ωt − π2( ) = VC cos ωt − π

2( )i.e., VC = IXC

=IωC

⎛ ⎝ ⎜

⎞ ⎠ ⎟ .

vC is 1

4 of a cycle ( 90!) behind i.

t vC(t)

90!

VC = I

ωC

I i(t)

Corresponding phasor diagram of the voltage and current for a capacitor.

" V C

vC

" I

ωt

i

Power in a capacitor in an a.c. circuit:The instantaneous power pC = vCi.

But i = Icos(ωt) and vC =

IωC

cos(ωt − 90!)

∴pC =

I2

ωCcos(ωt)cos(ωt − 90!)

The average power (over one cycle):

P av =

1T

I2

ωCcos(ωt)cos(ωt − 90!)

0

T∫ dt = 0.

... look at the graph!! ...The energy stored in the capacitor in one quarter-cycle is released in the next quarter-cycle.

t

t

I

VC

i

vC

pC

0 T

2 T

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Comparison of resistance and reactance ...

ω (= 2πf )

R

Resistance does not change with frequency

ω (= 2πf )

XL

Reactance of an inductor XL (= ωL) increases with frequency

ω (= 2πf )

Reactance of a capacitor

XC (=

1ωC)

decreases with frequency

XC

Question 29.4: Four resistors and a capacitor are connected as shown. What is the effective resistance of the circuit at very low frequency (ω→ 0), and very high frequency (ω→∞)?

~

10Ω 10Ω

20Ω 20Ω

0.01µF

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(a) At very low frequencies ( ω→ 0), XC =

1ωC

→ ∞.

(b) At very high frequencies (ω→∞), XC =

1ωC

→ 0.

10Ω

~~

10Ω

20Ω 20Ω

40Ω

10Ω

~~

10Ω

20Ω 20Ω

13.3Ω

1Req

=1

20Ω+

140Ω

=3

40Ω.

∴Req =

40Ω3

= 13.3Ω

By Kirchoff’s 2nd rule: the resultant p.d. supplied by the

generator is ! V =! V R + (

! V L +

! V C).

The angle δ between ! V and

! I is called the phase angle.

~L, vL

C, vC R, vR i

V

XL > XCp.d. LEADScurrent by δ

! V R

! V L

! V C

! V

ωt

! I

δ( ! V L +

! V C)

! V L

! V C

! V R

ωt ! V

( ! V L +

! V C)

! I

δ

XL < XCp.d. LAGS

current by δ

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! V R

! V L

! V C

! V

ωt

! I

δ( ! V L +

! V C)

XL > XC

V =! V = VR( )2 + VL − VC( )2

= (IR)2 + IXL − IXC( )2 = I R2 + XL − XC( )2 .

i.e., V = IZ (ac equivalent of Ohms’s law)where Z is the impedance, given by:

Z = R2 + ωL − 1ωC( )2

⇒ (units Ω).

Z⇒ Z(ω) is like the ac resistance of the circuit. The instantaneous emf supplied by the generator is

v = Vcos(ωt + δ) = IZcos(ωt + δ),

where δ = tan−1 VL − VC

VR

⎝ ⎜

⎠ ⎟ = tan−1 XL − XC

R⎛ ⎝ ⎜

⎞ ⎠ ⎟

= tan−1 ωL − 1

ωCR

⎝ ⎜ ⎜

⎠ ⎟ ⎟ .

! V R

! V L

! V C

! V

ωt

! I

δ( ! V L +

! V C)

XL > XC

The instantaneous power is:

P = vi = Vcos(ωt + δ) × Icos(ωt)

= VIcos2(ωt)cosδ + cos(ωt)sin(ωt)sin δ

= VIcos2(ωt)cosδ +

12sin(2ωt)sin δ.

If we average over one cycle we have, from earlier,

cos2(ωt) =

12

and sin(2ωt) = 0,

so second term → 0.

∴ P av =

12

VI cosδ

= VrmsIrms cosδ.

Note: VrmsIrms is the maximum possible power. power factor

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In summary ...

The amplitude of the emf supplied generator is: V = IZ

where V is the magnitude of the phasor:

! V =! V R +

! V L +

! V C

(not an algebraic sum!) and Z is the impedance, given by:

Z = R2 + ωL − 1ωC( )2 .

So, we have relationships between V, I and Z:

V = IZ, I =

VZ

and Z =

VI

... that are the “ac equivalent”of Ohm’s law.

~L, vL

C, vC R, vR i

V

What is the sigificance of resonance? At resonance,

ω = ω!, ∴Z = R,

i.e., the impedance is purely resistive.

Also ...

Since Z = R2 + ωL − 1ωC( )2 , then Z is a minimum at

some ω = ω! when ω!L =

1ω!C

, i.e., when XL = XC.

ω! =

1LC

is called the resonant frequency.

ω (= 2πf )

XL = ωL

XC = 1

ωC

R

ω!

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What’s the average power at resonance?

We know: P av = VrmsIrms cosδ

power factorwhere δ is phase angle between current and emf of the generator. But, at resonance XL = XC.

∴δ = 0,i.e., the current and p.d. are in-phase, so cosδ = 1.

So, at resonance

P(ω! ) av = VrmsIrms, i.e., a maximum.

Also, at resonance, Vrms = IrmsR.

∴ P(ω!) av = Vrms2

R = Irms2R.

" V L

" V C

" V

ωt XL = XC

" V L −

" V C = 0

" I

Phasor diagramat resonance

How does the current vary with frequency? Take a generator with a constant voltage amplitude.

We have:

I =VZ

=V

R2 + ωL − 1ωC( )2

=V

R2 + L2 ω − 1ωLC( )2 =

V

R2 + L2 ω − ω!2ω

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2 ,

i.e.,

I =Vω

ω2R2 + L2 ω2 − ω!2( )2 .

Note, when ω = ω!, I =

VR

i.e., pure resistance.

So, at resonance, the current is determined only by R (and it is a maximum ).

~L

CRI

V

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ω! =

1LC

=1

(2.0 H)(0.5×10−6F)= 1000 rad/s,

and independent of R.

∴f! =

ω!2π

= 159.2Hz.

Note: Imax =

VR

, when ω = ω!.

What does this function look like?

I =Vω

ω2R2 + L2 ω2 − ω!2( )2

L = 2.0 H

C = 0.5 µF

V = 100 V

500 1000 1500 rad/s ω (= 2πf )

1.0

0.2

0.4

0.6

0.8

100Ω

200Ω

500Ω

1000Ω

ω!

ω! I(A)

How does the phase angle vary with frequency?

! V R

! V L

! V C

! V

ωt

! I

δ( ! V L −

! V C)

XL > XC

p.d. leadscurrent

XL > XC

Current leadsp.d.

XL < XC

0

500 1000 1500 rad/s

100Ω 500Ω

1000Ω

δ = tan−1 ωL − 1

ωCR

⎝ ⎜ ⎜

⎠ ⎟ ⎟

L = 2.0 H: C = 0.5 µF

δ (rad)

ω (= 2πf )

ω"

ω" π

2

π

4

−π

4

−π

2

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Q-value of a resonant circuit is defined as

Q =

ω!LR

≈ω!Δω

=f!Δf

where Δω ( Δf ) is the width of the power peak at half-maximum.

Larger Q ⇒ sharper peak.

How does the power vary with frequency?

P av = Irms2R =

Vrms2ω2R

ω2R2 + L2 ω2 − ω!2( )2.

Maximum power: Pav(ω = ω!)⇒Vrms

2R .

L = 2.0 H

C = 0.5 µF

V = 100 V

Vrms = 70.7 V

500 1000 1500 rad/s

ω (= 2πf )

100Ω

500Ω

Δω

ω!

P av 500W

100W

Question 29.5: If the frequency difference between radio stations on the old AM band is 10kHz, what are suitable Q values at the lowest AM frequency (500kHz) and the highest AM frequency (1600kHz)?

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The power curves for two neighboring stations ( fi and f j)

should be well separated so that only one of them is “detected” at a time.

Make the width of the power functions ( Δf ) about 1

5th

(i.e., 20%) of the channel separation, i.e., Δf ~ 0.2(f j − fi ) = 2kHz .

At the low frequency end ( f = 500kHz):

Q =

fΔf

=500kHz2kHz

= 250.

At the high frequency end ( f = 1600kHz):

Q =

fΔf

=1600kHz

2kHz= 800.

Therefore, we need a Q value > 800.

Δf

fi f j

If f j − fi = 10kHz then

Δf << 10kHzfor “good” separation.

Question 29.6: A series RLC circuit, with L = 10 mH,

R = 5.0 Ω and C = 2.0 µF, is driven by an ac voltage source that has a peak emf of 100 V. Find

(a) the resonant frequency, and(b) Irms at resonance.

When the frequency is 8000 rad/s, find(c) the reactances of the capacitor and the inductor,(d) the impedance of the circuit, and(e) the value of Irms, and(f) the phase angle.

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Given : L = 10 mH,

C = 2.0 µF, R = 5.0 Ω and

Vpeak = 100 V.

(a) ω! = 1

LC = 1(10 ×10−3H)(2.0 ×10−6F)

= 7.07 ×103 rad/s (1125 Hz).

(b) Irms =Vrms

Z. At resonance Z⇒ R = 5Ω.

Vrms =Vpeak

2 = 100V2 = 70.7 V.

∴Irms = 70.7 V5.0Ω = 14.1 A.

(c) XC = 1ωC and XL = ωL, where ω = 8000 rad/s.

∴XC = 1

(8000 rad/s)(2.0 ×10−6F)= 62.5Ω,

and XL = (8000 rad/s)(10 ×10−3H) = 80.0Ω.

~L

CRI

V Given : L = 10 mH,

C = 2.0 µF, R = 5.0 Ω and

Vpeak = 100 V.

(d) Z = R2 + ωL − 1ωC( )2

= R2 + XL − XC( )2

= (5.0Ω)2 + 80Ω − 62.5Ω( )2 = 18.2Ω.

(e) Irms = VrmsZ = 70.7 V

18.2Ω = 3.88 A.

(f) δ = tan−1 XL − XC

R⎛ ⎝

⎞ ⎠ = tan−1 80.0Ω− 62.5Ω

5.0Ω⎛ ⎝

⎞ ⎠

= 74.1!.

~L

CRI

V

" V R

" V L

" V C

" V

ωt

" I

δ( " V L −

" V C)

XL > XC

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Question 29.7: For the circuit described in the previous question (29.6), find

(a) the Q factor, and(b) the resonance width (in Hz).(c) What is the power factor when ω = 8000 rad/s?

(a) The Q factor is defined as Q =

ω!LR

=ω!LR

=(7.07 ×103 rad/s)(10 ×10−3H)

5.0Ω= 14.1,

using ω! from the previous question.

(b) From earlier, Q =

ω!Δω

=f!Δf

, i.e., Δf =

f!Q

=ω!

2πQ

=

7.07 ×103 rad/s2π ×14.1

= 79.8 Hz.

(c) The power factor is cosδ. In the previous problem

δ = tan−1 ωL − 1

ωCR

⎝ ⎜

⎠ ⎟ = 74.1!.

∴cosδ = 0.27.

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Question 29.8: A certain electrical device draws a rms current of 10 A at an average power of 720 W when connected to a 120 V (rms), 60 Hz power source.

(a) What is the impedance of the device?(b) What series combination of resistance and reactance would have the same impedance as the device?(c) If the current leads the emf, is the reactance inductive or capacitive?

Irms = 10 A: P av = 720 W

Vrms = 120 V.

(a) By definition: Vrms = IrmsZ.

∴Z = Vrms

Irms= 120 V

10 A = 12 Ω.

(b) Also, P av = Irms2R, where R is the resistance of the

“device” ... why only R??

∴R = P av

Irms2 = 720 W

(10 A)2 = 7.2 Ω.

From earlier: Z = R2 + (XL − XC)2 ,

∴Z2 = R2 + (XL −XC )2 = R2 + X2,

where X is the reactance of the ‘device’.

∴X = Z2 − R2 = (12Ω)2 − (7.2Ω)2 = 9.6 Ω.

(c) If the current leads the emf the reactance is capacitive.

~V

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Question 29.9: So far, we have considered only ideal inductors and ideal capacitors, i.e., ones in which there is no resistance. What happens to the voltage and current with a non-ideal inductor, i.e., one with finite resistance?

If the inductor has resistance, it is equivalent to an ‘ideal’ inductor with a resistance in series. Construct the corresponding phasor diagram: the voltage across the ‘ideal’ inductor (with R = 0) leads the current by

90! but in a resistor and inductor series circuit the

phase angle between voltage and the current is

0 < δ < 90!.

In fact,

δ = tan−1 VL

VR

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = tan−1 ωL

R⎛ ⎝

⎞ ⎠ .

~ L R

" V R

" V

" I

" V L δ

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By Faraday’s law, in the primary coil:

ε1 = −N1

dΦdt

i.e.,

dΦdt

= −ε1N1

.

In the secondary: ε2 = −N2

dΦdt

. But

dΦdt

⇒ −ε1N1

.

ε1ε2

=N1N2

• If ε2 > ε1 STEP-UP TRANSFORMER• If ε2 < ε1 STEP-DOWN TRANSFORMER

Primary coil

Secondary coil

Magnetic (iron) core(laminated)

N2 ⇒ Number of turns on the secondary coil

ε1

ε2

N1 ⇒ Number of turns on the primary coil

Large, well designed transformers convert up to 99.5% of the input power into output power.

Losses occur through:

HEAT and VIBRATION (SOUND)

POWER IN ~ POWER OUT

i.e., ε1 i1 ≈ ε2 i2

Power in Power out TRANSFORMER

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Question 29.10: A step-up transformer operates on a 110 V source and supplies a resistive load with 2.00 A. The ratio of primary and seconday windings is 1:25.

(a) What is the secondary voltage?(b) What is the current in the primary?(c) What is the power output?

Assume 100% efficiency. All voltages and currents are rms values.

(a) We have

V1,rmsV2,rms

=N1N2

, i.e., V2,rms =

N2N1

V1,rms

= 25 ×110 V = 2750 V (rms).

(b) With no power loss V1,rmsI1,rms = V2,rmsI2,rms.

∴I1,rms =

V2,rmsI2,rmsV1,rms

=2750 V × 2 A

110 V= 50.0 A (rms).

(c) With no losses P2 = P1 = V1,rmsI1,rms

= (110 V)(50.0 A) = 5500 W.

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Typical step-up transformers at a hydro-electric power plant.

5,000 VPower Station

10,000 Vlocal distribution

Transmission lines 250,000 V

Home120V

M Area sub-station

25 : 1Utility pole

83:1

Power station 1 : 50

10’s to 100’s miles250,000 V

120 V 10,000 V

Home

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Typical step-down transformers used to produce domestic voltage (110V/120V); they are often located on local neighborhood utility poles.

Question 29.11: You return from a trip to London with an appliance advertized as the world’s greatest coffee maker. Unfortunately, it was designed to operate from a 240 V source to obtain the 960 W of power that it requires.

(a) What can you do to operate it at 120 V?(b) What current will the coffee maker draw from the 120 V line?(c) What is the resistance of the coffee maker?

All voltages are rms values.

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(a) To obtain V2 = 240 V with V1 = 120 V you need a

step-up transformer with a turns ratio

N2N1

=V2V1

=240120

= 2,

i.e., the secondary of the transformer needs twice as many turns as the primary.

(b) Assuming no losses P1 = P2 = 960 W.

But P1 = V1I1, where I1 is the rms current in the 120 V

primary.

∴I1 =

P1V1

=960 W120 V

= 8.0 A.

(c) The rms current in the secondary is

I2 =

P2V2

=960 W240 V

= 4.0 A.

∴R =

V2I2

=2404.0

= 60 Ω.

Therefore, the transformer must be capable of carrying a current of 8.0 A in the primary circuit and 4.0 A in the secondary circuit.