Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 29, Problem 69P

(a)

To determine

RMS voltage between the points A and B.

(a)

Expert Solution
Check Mark

Answer to Problem 69P

  VAB=80.344V

Explanation of Solution

Given:

Rms voltage of the generator, ε=115V

Frequency of the generator, f=60Hz

Inductance, L=137mH

Resistance, R=50Ω

Capacitance, C=25μF

Formula used:

  Irms=εZZ=R2+ ( X L X C )2XL=2πfLXC=12πfC

Calculation:

The rms voltage between points A and B is,

  VAB=IrmsXLVAB=εZXLVAB=ε R 2 + ( X L X C ) 2 XLVAB=ε R 2 + ( 2πfL 1 2πfC ) 2 2πfLVAB=115V ( 50Ω ) 2 + ( ( 2π×60Hz×137mH ) 1 ( 2π×60Hz×25μF ) ) 2 (2π×60Hz×137mH)VAB=115V ( 50Ω ) 2 + ( 51.648Ω106.10Ω ) 2 51.648ΩVAB=80.344V

Conclusion:

The rms voltage through points A and B where the inductor is present is, VAB=80.344V .

(b)

To determine

RMS voltage between B and C.

(b)

Expert Solution
Check Mark

Answer to Problem 69P

  VBC=77.78V

Explanation of Solution

Given:

Rms voltage of the generator, ε=115V

Frequency of the generator, f=60Hz

Inductance, L=137mH

Resistance, R=50Ω

Capacitance, C=25μF

Formula used:

  Irms=εZZ=R2+ ( X L X C )2XL=2πfLXC=12πfC

Calculation:

The rms voltage between points B and C is,

  VBC=IrmsRVBC=εZRVBC=ε R 2 + ( X L X C ) 2 RVBC=ε R 2 + ( 2πfL 1 2πfC ) 2 RVBC=115V ( 50Ω ) 2 + ( ( 2π×60Hz×137mH ) 1 ( 2π×60Hz×25μF ) ) 2 50ΩVBC=115V ( 50Ω ) 2 + ( 51.648Ω106.10Ω ) 2 50ΩVBC=77.78V

Conclusion:

Across the resistor in between the points B and C, the rms voltage is VBC=77.78V .

(c)

To determine

RMS voltage between points C and D.

(c)

Expert Solution
Check Mark

Answer to Problem 69P

  VCD=165.05V

Explanation of Solution

Given:

Rms voltage of the generator, ε=115V

Frequency of the generator, f=60Hz

Inductance, L=137mH

Resistance, R=50Ω

Capacitance, C=25μF

Formula used:

  Irms=εZZ=R2+ ( X L X C )2XL=2πfLXC=12πfC

Calculation:

The rms voltage between points D and C is

  VCD=IrmsXCVCD=εZXCVCD=ε R 2 + ( X L X C ) 2 XCVCD=ε R 2 + ( 2πfL 1 2πfC ) 2 XCVCD=115V ( 50Ω ) 2 + ( ( 2π×60Hz×137mH ) 1 ( 2π×60Hz×25μF ) ) 2 2π×60Hz×25μFVCD=115V ( 50Ω ) 2 + ( 51.648Ω106.10Ω ) 2 106.10ΩVCD=165.05V

Conclusion:

Rms voltage across the capacitance is VCD=165.05V .

(d)

To determine

The rms voltage between the points A and C.

(d)

Expert Solution
Check Mark

Answer to Problem 69P

  VAC=111.58V

Explanation of Solution

Given:

Rms voltage of the generator, ε=115V

Frequency of the generator, f=60Hz

Inductance, L=137mH

Resistance, R=50Ω

Capacitance, C=25μF

Rms voltage between AB, VAB=80.344V

Rms voltage between BC, VBC=77.78V

Calculation:

The voltage across AC is calculated using the Pythagoras theorem as the voltage in the inductor leads the voltage in the resistor according to the phasor diagrams,

  VAC=V AB2+V BC2=80V2+77.78V2=111.58V

Conclusion:

The voltage across AC is VAC=111.58V .

(e)

To determine

The rms voltage across BD.

(e)

Expert Solution
Check Mark

Answer to Problem 69P

  VBD=182.46V

Explanation of Solution

Given:

Rms voltage of the generator, ε=115V

Frequency of the generator, f=60Hz

Inductance, L=137mH

Resistance, R=50Ω

Capacitance, C=25μF

Rms voltage between AB, VAB=80.344V

Rms voltage between DC, VCD=165.05V

Calculation:

The voltage across CD is such that the voltage across the resistor leads the capacitor voltage according to the phasor diagram,

  VBD=V CD2+V BC2VBD=165 .05V2+77.78V2VBD=182.46V

Conclusion:

Across BD the rms voltage is VBD=182.46V .

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