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Holt Physics
Problem 22D
TRANSFORMERS
PROBLEM
An ac generator at a central power station uses a step-up transformer to
provide a potential difference of 68 kV across the secondary coil. If the
primary coil has 125 turns and the secondary coil has 625 turns, what is
the potential difference across the primary coil?
SOLUTION
Given:
∆V2 = 68 kV = 6.8 × 104 V
N2 = 625 turns
Unknown:
∆V1 = ?
N1 = 125 turns
Choose the equation(s) or situation: Use the transformer equation.
∆V1 N1
 = 
∆V2 N2
Rearrange the equation(s) to isolate the unknown(s): Rearrange the equation
for a transformer to solve for the potential difference in the primary coil.
∆V N
∆V1 = 21
N2
Substitute the values into the equation(s) and solve:
∆V N
(6.8 × 104 V)(125)
∆V1 = 21 =  = 1.4 × 104 V = 14 kV
N2
(625)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Evaluate: The potential difference across the primary should be 14 kV. The stepup factor for the transformer is 1:5.
ADDITIONAL PRACTICE
1. A transmission line to a city has a potential difference of 6.9 kV across
the secondary coil. If the primary coil has 1400 turns and the secondary
coil has 140 turns, what is the potential difference across the primary
coil?
2. A power line has a potential difference of 3.4 kV across the secondary
coil. If the primary coil has 9.0 × 101 turns and the secondary coil has
2250 turns, what is the potential difference across the primary coil?
3. A transmission line to a city has a potential difference of 46 kV across the
primary coil. If the primary coil has 1250 turns and the secondary coil
has 250 turns, what is the potential difference across the secondary coil?
4. A high-voltage cable has a potential difference of 5.6 kV across the primary coil. If the primary coil has 140 turns and the secondary coil has
840 turns, what is the potential difference across the secondary coil?
Problem 22D
Ch. 22–7
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5. A high-voltage cable has a potential difference of 9.2 kV across the primary coil. If the primary coil has 120 turns and the secondary coil has
1200 turns, what is the potential difference across the secondary coil?
6. A ac power station has a potential difference of 36 kV across the primary
coil and a potential difference of 7.2 kV across the secondary coil. If the
primary coil has 55 turns, how many turns does the secondary coil have?
7. The plug to your zip drive has a step-down transformer in it. The potential difference across the primary coil is 240 V and a potential difference
of 5.0 V across the secondary. What is the step-down ratio? (Hint: the
step-down ratio is the ratio of N1:N2.)
8. A transmission line has a potential difference of 3.6 kV across the primary coil and a potential difference of 1.8 kV across the secondary coil. If
the primary coil has 58 turns, how many turns does the secondary coil
have?
9. A central ac power station has a potential difference of 49 kV across the
primary coil and a potential difference of 4.9 kV across the secondary
coil. If the secondary coil has 480 turns, how many turns does the primary coil have?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
10. A ac generator central power station can produce 1380 kW of power. The
secondary coil has a potential difference of 3.4 kV. If the primary coil has
340 turns, and the secondary coil has 17 turns, what is the potential difference across the primary coil? What is the current in the primary coil?
Ch. 22–8
Holt Physics Problem Bank
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Givens
Solutions
10. Irms = 2.2 × 1010 A
I s
2.2 × 1010 A
Imax = rm
=  = 3.1 × 1010 A
0.707
0.707
R = 6.1 × 10−10 Ω
P = (Irms)2 R = (2.2 × 1010 A)2(6.1 × 10−10 Ω) = 2.9 × 1011 W
Additional Practice 22D
1. ∆V2 = 6.9 × 103 V
N1 = 1400 turns
N
1400
∆V1 = ∆V2 1 = (6.9 × 103 V)  = 6.9 × 104 V
N2
140
N2 = 140 turns
2. ∆V2 = 3.4 × 103 V
N1 = 90 turns
90
N
∆V1 = ∆V2 1 = (3.4 × 103 V)  = 1.4 × 102 V
2250
N2
N2 = 2250 turns
3. ∆V1 = 4.6 × 104 V
N1 = 1250 turns
N
250
∆V2 = ∆V1 2 = (4.6 × 104 V)  = 9.2 × 103 V
N1
1250
N2 = 250 turns
N
840
∆V2 = ∆V1 2 = (5600 V)  = 3.36 × 104 V
N1
140
4. ∆V1 = 5600 V
N1 = 140 turns
N2 = 840 turns
N1 = 120 turns
N2 = 1200 turns
∆V2 = 7.2 × 103 V
N1 = 55 turns
7. ∆V1 = 240 V
N
1200
∆V2 = ∆V1 2 = (9200 V)  = 9.20 × 104 V
N1
120
6. ∆V1 = 3.6 × 104 V
∆V
7.2 × 103 V
= 11 turns
N2 = N1 2 = (55) 
∆V1
3.6 × 104 V
N1 ∆V1 240 V
 =  =  = 48:1
N2 ∆V2 5.0 V
∆V2 = 5.0 V
∆V
3600 V
N2 = N1 2 = (58)  = 1.2 × 102 turns
∆V1
1800 V
8. ∆V1 = 1800 V
∆V2 = 3600 V
N1 = 58 turns
9. ∆V1 = 4900 V
4
∆V2 = 4.9 × 10 V
∆V
4900 V
N1 = N2 1 = (480) 
= 48 turns
∆V2
4.9 × 104 V
N2 = 480 turns
10. P = 1.38 × 106 W
3
V
∆V2 = 3.4 × 10 V
N1 = 340 turns
N2 = 17 turns
V Ch. 22–5
N
340
∆V1 = ∆V2 1 = (3.4 × 103 V)  = 6.8 × 104 V
N2
17
P = ∆V1I1
P
1.38 × 106 W
I1 =  = 
= 2.0 × 101 A
∆V1 6.8 × 104 V
Holt Physics Solutions Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
5. ∆V1 = 9200 V