Φ21 Fall 2006 HW18 Solutions

1 Transformers

Figure 1: Sketch of a transformerwith the primary coil attached toan AC source and the secondaryattached to a resistor.

Learning Goal: To understand the concepts explaining the operationof transformers. (Note, the Printable Solutions on MasteringPhysicshave a lot of information about transformers in the introduction to thisproblem.) The ratio of the EMF's in the primary and secondary is thesame as the ratio of the number of turns. The EMF in an ideal coil isalso the voltage across it.

E2

E1=

V2

V1=

N2

N1

Also, the transformer doesn't add or take away energy, so the power inthe primary and secondary is equal:

V1I1 = V2I2

Part A. The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primarycoil is connected to a size AA battery that supplies a constant voltage of 1.5 volts. What voltage would bemeasured across the secondary coil?

Solution: There would be no voltage across the secondary. A DC current does not cause an EMF.

Part B. A transformer is intended to decrease the RMS value of the alternating voltage from 500 volts to25 volts. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.

Solution: To step the voltage down from 500 V to 25 V, the ratio of the number of turns would be 500:25= 20:1. Thus, N2 is:

N2 =E2

E1N1 =

25V500 V

200 = 10

Part C. A transformer is intended to decrease the rms value of the alternating current from 500 amperes to25 amperes. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondarycoil.

Solution: Since the power going in equals the power going out, the ratio of the currents is the inverse ofthe ratio of the voltages.

N2 =I1

I2N1 =

500A25 A

200 = 4000

Part D. In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns. Ifthe primary current is 2.5 amperes, what is the secondary current I2?

Solution: This is a step-down transformer with a ratio of 5:1. Assuming the primary current is AC, thesecondary current would be

I2 =N1

N2I1 =

40080

2.5A = 12.5 A

Part E. The primary coil of a transformer has 200 turns and the secondary coil has 800 turns. The powersupplied to the primary coil is 400 watts. What is the power generated in the secondary coil if it is terminatedby a 20-ohm resistor?

Solution: Since the power in the primary and secondary is the same, the power supplied to the resistor willbe 400 W.

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Part F. The primary coil of a transformer has 200 turns, and the secondary coil has 800 turns. Thetransformer is connected to a 120-volt (rms) ac source. What is the (rms) current I1 in the primary coil ifthe secondary coil is terminated by a 20-ohm resistor?

Solution: The RMS voltage in the secondary will beV2 = (N2/N1) V1 = (800/200) · 120 = 480V

With the 20 Ω resistor, the current in the secondary will be I2 = V2/R = (480V) / (20Ω) = 24A.

In order to produce this secondary current, the primary current would have to be larger

I1 =N2

N1I2 =

800200

24A = 96A

Part G. A transformer supplies 60 watts of power to a device that is rated at 20 volts (rms). The device isconnected to the secondary coil. The primary coil is connected to a 120-volt (rms) ac source. What is thecurrent I1 in the primary coil?

Solution: The power is the same on both sides of the transformer, so the current is

I1 =P

V1=

60W120 V

= 0.5A

Part H. The voltage and the current in the primary coil of a nonideal transformer are 120 V and 2.0 A.The voltage and the current in the secondary coil are 19.4 V and 11.8 A. What is the eciency e of thetransformer? The eciency of a transformer is dened as the ratio of the output power to the input power,expressed as a percentage: e = 100% (Pout/Pin).

Solution: The eciency is

e = 100%(19.4V) (11.8A)(120V) (2.0A)

= 95.4%

2 Problem K34.19

A square parallel-plate capacitor 5.30 cm on a side has a 0.540 mm gap. What is the displacement current

in the capacitor if the potential dierence across the capacitor is increasing at 500,000 V/s? (Displacementcurrent is dened on page 1100 of the book. It is part of the extra term in Ampere's Law.)

Solution: The displacement current is dened as ID = ε0dΦE

dt , where ΦE =∫

~E ·d ~A is the electric ux. Just

like magnetic ux, usually ΦE = EA. Taking the time derivative, ID = ε0AdEdt . The area is A = (0.0530m)2

and the electric eld and voltage in the capacitor are related by V = Ed, so

ID = ε0A

d

dV

dt= ε0

(0.0530m)2

(0.000540m)(500, 000 V/s) = 2.30 × 10−5 A

3 Problem K35.23

What capacitor in series with a 100 Ω resistor and a 23.0 mH inductor will give a resonance frequency of1010 Hz?

Solution: The resonant frequency depends only on the inductor and the capacitor.

ω2 = (2πf)2 =1

LC⇒ C =

1(2πf)2 L

=1

(2π1010Hz)2 (23 × 10−3 H)= 1.08 × 10−6 F

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4 Problem K35.48

A series RLC circuit consists of a 100 Ω resistor, a 0.10 H inductor, and a 100 µF capacitor. It is attachedto a 120 V/60 Hz power line.

Part A. What is the peak current I0?

Figure 2: Phasor diagram for anRLC circuit.

Solution: The loop equation for a series RLC circuit is

E(t) − I(t)R − LdI

dt− 1

CQ(t) = 0

Where the charge on the capacitor is Q(t) =∫

I(t) dt with the arbitraryconstant set so the average Q is zero. The current as a function of timeis

I(t) = I0 cos ωt + φ

Because of the sin's and cos's, to nd I0, we draw the phasor diagram. The lengths of the vectors are

VR = I0R

VL = I0XL = I0ωL

VC = I0XC =I0

ωC

E0 = VT =√

V 2R + (VL − VC)2

The reactances are XL = ωL = 2π (60Hz) (0.1H) = 37.7Ω and XC = 1ωC = 1

2π(60 Hz)(100×10−6 F) = 26.5Ω.

By solving for I0, noting that E0 =√

2VRMS, we get

I0 =E0√

R2 + (XL − XC)2=

E0√R2 +

(ωL − 1

ωC

)2=

√2 120V√

(100Ω)2 + (37.7Ω − 26.5Ω)2= 1.69A

Part B. What is the phase angle φ in degrees?

Solution: The phase angle is the angle between the total voltage VT and the total current I0 in the phasordiagram. By inspection, it has the value

φ = tan−1

(VL − VC

VR

)= tan−1

(XL − XC

R

)= tan−1

(37.7Ω − 26.5Ω

100Ω

)= 0.111 radians = 6.39

Part C. What is the avergae power loss?

Solution: The only component with a power loss is the resistor. The instantaneous power is P = IV , andthe average power is

Pave = IRMSVRMS = (IRMS)2 R =I20R

2=

(1.69A)2 (100 Ω)2

= 142W

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