Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 18, Problem 127P
To determine

Find the voltmeter reading.

Expert Solution & Answer
Check Mark

Answer to Problem 127P

The voltmeter reading is 7.22 V.

Explanation of Solution

The circuit diagram is shown below,

Physics, Chapter 18, Problem 127P

From the circuit the voltmeter reading is,

V4=I4(83.0 Ω) (I)

From Kirchhoff’s junction rule,

I1=I2+I3 (II)

I3=I4+I5 (III)

Write the equation to find I1 .

I1=εReq (IV)

Here, I1 is the current, ε is potential difference and Req is the equivalent resistance.

From the circuit, consider 35 Ω resistor as R1, 1.40 kΩ resistor as R2, 16 kΩ resistor as R3, 83 kΩ resistor as R4 and voltmeter resistance 670 kΩ as R5.

The resistors R4 and R5 are in parallel then the equivalent resistance is,

1R'=1R4+1R5R'=(1R4+1R5)1

This resistance is in series with R3 , the equivalent resistance is,

R''=R3+R'

This resistance is in parallel with R2 then the equivalent resistance is,

1R'''=1R2+1R''R'''=11R2+1R''

This resistance is in series with R1, then the final equivalent resistance is,

Req=R1+R'''

Substitute the value of R', R'' and R''' in the above equation.

Req=R1+11R2+1R3+(1R4+1R5)1 (V)

Conclusion:

Substitute 35 Ω for R1, 1.40 kΩ for R2, 16 kΩ for R3, 83 kΩ for R4 and 670 kΩ for R5 in equation V.

Req=35 Ω+111.40 kΩ+116 kΩ+(183 kΩ+1670 kΩ)1=35 Ω+111.40 kΩ+116 kΩ+73851 Ω=35 Ω+111.40 kΩ+189851 Ω=35 Ω+1379 Ω

Substitute 35 Ω+1379 Ω for Req and 9.00 V for ε in equation IV.

I1=9.00 V35 Ω+1379 Ω=6.367×103 A

Then the voltage across the all the resistors drawn vertical in the circuit with equivalent resistance of 1379 Ω is (6.367×103 A)(1379 Ω)=8.780 V.

This voltage is across the two resistors in right side with equivalent resistance 89851 Ω and the voltmeter meter. Then the current I3 is 8.780 V89851 Ω=97.72 μA.

The voltage across the voltmeter and the resistor it is supposed to be measured is then (97.72 μA)(73851 Ω)=7.22 V.

Therefore, the voltmeter reading is 7.22 V.

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Chapter 18 Solutions

Physics

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