Physical Science - With Lab Manual
Physical Science - With Lab Manual
11th Edition
ISBN: 9781260021417
Author: Tillery
Publisher: MCG
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Textbook Question
Chapter 6, Problem 15PEB

The step-down transformer in a local neighborhood reduces the voltage from a 7,200 V line to 120 V. (a) If there are 125 loops on the secondary, how many are on the primary coil? (b) What current does the transformer draw from the line if the current in the secondary is 36 A? (c) What are the power input and output?

(a)

Expert Solution
Check Mark
To determine

The number of loops at the primary side of a step-down transformer.

Answer to Problem 15PEB

Solution:

The number of loops at the primary side is 7500.

Explanation of Solution

Given data:

A step-down transformer with 125 loops on the secondary side. The voltage at the primary side is 7200 V and the voltage at the secondary side is 120 V. The current on secondary side is 36 A.

Formula used:

The expression that relates the number of loops and the voltage is written as,

VpNp=VsNs

Here, Vp is the primary voltage, Np are the number of loops at the primary side, Vs is the secondary voltage and Ns are the number of loops at the secondary side.

Also, power and current can be determined by the formula, therefore

VpIp=VsIs

Here, Ip and Is are current at primary and secondary side, respectively.

Explanation:

Consider the expression,

VpNp=VsNs

Rearrange to get the secondary voltage Vs.

Np=NsVpVs

Substitute 125 for Ns, 7200 V for Vp and 120 V for Vs.

Np=NsVpVs=125×7200 V120 V=7500

Conclusion:

Hence, the number of loops ate the primary side is 7500.

(b)

Expert Solution
Check Mark
To determine

The primary current in a step down transformer.

Answer to Problem 15PEB

Solution:

The primary current is 0.6 A.

Explanation of Solution

Given data:

A step-down transformer with 125 loops on the secondary side. The voltage at the primary side is 7200 V and the voltage at the secondary side is 120 V. The current on secondary side is 36 A.

Formula used:

The expression for power in a transformer is written as,

VpIp=VsIs

Here, Ip and  Is are current at primary and secondary side, respectively.

Explanation:

Consider the expression for power in the transformer.

VpIp=VsIs

Rearrange the power expression to get the primary current Ip.

Ip=VsIsVp

Substitute 120 V for Vs, 36 A for Is and 7200 V for Vp.

Ip=VsIsVp=120 V×36 A7200 V=0.6 A

Conclusion:

Hence, the primary current is 0.6 A.

(c)

Expert Solution
Check Mark
To determine

The power of the transformer at the input and the output sides.

Answer to Problem 15PEB

Solution: The power at the input and the output side is same and is equal to 4320 W.

Explanation of Solution

Given data:

A step-down transformer with 125 loops on the secondary side. The voltage at the primary side is 7200 V and the voltage at the secondary side is 120 V. The current on secondary side is 36 A.

Formula used:

The expression for power at the input side is written as,

Pi=VpIp

Here, Pi is the input power, Vp is the primary voltage and Ip is the primary current.

The expression for power at the output side is written as,

Po=VsIs

Here, Po is the output power, Vs is the secondary voltage and Is is the secondary current.

Explanation:

Consider the expression for the input power.

Pi=VpIp

Substitute 7200 V for Vp and 0.6 A for Ip.

Pi=VpIp=7200 V×0.6 A=4320 W

Similarly, consider the expression for the output power.

Po=VsIs

Substitute 120 V for Vs and 36 A for Is.

Po=VsIs=120 V×36 A=4320 W

Conclusion:

Hence, the power at the input and the output side is same and is equal to 4320 W.

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Chapter 6 Solutions

Physical Science - With Lab Manual

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