COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
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Chapter 21, Problem 46QAP
To determine

(a)

The ratio of number of turns in the primary and the secondary coil.

Expert Solution
Check Mark

Answer to Problem 46QAP

The ratio of number of turns in secondary and the primary coil is 41.

Explanation of Solution

Given info:

The primary voltage is 120Vrms.

The secondary voltage is 5000Vrms.

The power dissipated value is 75W.

Formula used:

The formula to find the ratio of number of turns in secondary and the primary coil is,

  NsNp=VsVp   Equation-1

Here, Vp is the primary voltage and Ns is the number of turns in the secondary coil and Np is the number of turns in the primary coil.

Calculation:

Substitute the given values in equation-1 to find the ratio.

  NsNp=5000Vrms120Vrms=41 Equation-2

Conclusion:

From equation -2 the ratio of number of turns in secondary and the primary coil is 41.

To determine

(b)

The value of the rms current in the primary and the secondary coil.

Expert Solution
Check Mark

Answer to Problem 46QAP

The value of the rms current in the secondary coil is 0.015A and the primary coil is 0.625A.

Explanation of Solution

Given info:

The primary voltage is 120Vrms.

The secondary voltage is 5000Vrms.

The power dissipated value is 75W.

Formula used:

The formula to find the rms current is,

  Irms=PVrms   Equation-3

Here, Vrms is the rms voltage and P is the power dissipated.

Calculation:

Substitute the given values in equation-1 to find the rms current in the primary and secondary coil.

Primary coil rms current Ip(rms) is,

  Ip(rms)=75W120Vrms=0.625A Equation-4

Secondary coil rms current Is(rms) is,

  Is(rms)=75W5000Vrms=0.015A Equation-5

Conclusion:

From equation -4 and equation-5 the rms current in the secondary coil is 0.015A and the primary coil is 0.625A.

To determine

(c)

The the effective resistance of the primary coil.

Expert Solution
Check Mark

Answer to Problem 46QAP

The effective resistance of the primary coil is 192Ω.

Explanation of Solution

Given info:

The primary voltage is 120Vrms.

The secondary voltage is 5000Vrms.

The power dissipated value is 75W.

Formula used:

The formula to find the effective resistance of the primary coil is,

  Rp=Vp(rms)2P   Equation-3

Here, Vp(rms) is the rms voltage of the primary coil and P is the power dissipated.

Calculation:

Substitute the given values in equation-1 to find the effective resistance.

Primary coil rms current Ip(rms) is,

  Rp=( 120 V rms )275W =192Ω Equation-6

Conclusion:

From equation-6 effective resistance of the primary coil is 192Ω.

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Chapter 21 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

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