Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Question
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Chapter 19, Problem 77GP
To determine

The possible values of the terminal voltage Vba in the given circuit.

Expert Solution & Answer
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Answer to Problem 77GP

The possible values of the terminal voltage Vba are 52.3V and 28.3V .

Explanation of Solution

Given information:

Consider the given circuit.

Physics: Principles with Applications, Chapter 19, Problem 77GP

Formula used:

Apply Kirchhoff’s voltage law in the circuit to determine the currents.

Calculation:

Since, the current across 4kΩ resistor is 3.50mA . Calculate the voltage across nodes c and d using Ohm’s law.

  Vcd=4×103×3.50×103=14V

  4kΩ resistor and 8kΩ are connected in parallel. So, the voltage across the node will be same.Let Ief is the current through 8kΩ resistor. Calculate the value of the current using Ohm’s law.

  Ief=148×103=1.75mA

Calculate the total current flowing through 5 resistor.

  I=1.75+3.50=5.25mA

Since the value of the voltage Vba is not given. Calculate the value of the terminal voltage Vba when current flowing towards 5 resistor using Kirchhoff’s voltage law.

  Vba+(5)I+14+12+1I=0Vba+(5001)I+26=0

… (1)

Substitute the value of the current in equation (1).

  Vba+(5001)(5.25×103)+26=0Vba=52.3V

Calculate the value of the terminal voltage Vba when current flowing away from 5 resistor using Kirchhoff’s voltage law.

  Vba+(5)I+1412+1I=0Vba=(5001)I2=(5001)×5.25×1032=28.3V

Therefore, the possible values of the terminal voltage Vba are 52.3V and 28.3V .

Chapter 19 Solutions

Physics: Principles with Applications

Ch. 19 - Prob. 11QCh. 19 - Prob. 12QCh. 19 - Prob. 13QCh. 19 - Prob. 14QCh. 19 - Prob. 15QCh. 19 - Prob. 16QCh. 19 - Given the circuit shown in Fig. 19-38, use the...Ch. 19 - Prob. 18QCh. 19 - Prob. 19QCh. 19 - 19. What is the main difference between an analog...Ch. 19 - What would happen if you mistakenly used an...Ch. 19 - Prob. 22QCh. 19 - Prob. 23QCh. 19 - Prob. 24QCh. 19 - Prob. 1PCh. 19 - Prob. 2PCh. 19 - Prob. 3PCh. 19 - Prob. 4PCh. 19 - Prob. 5PCh. 19 - Prob. 6PCh. 19 - Prob. 7PCh. 19 - Prob. 8PCh. 19 - Prob. 9PCh. 19 - Prob. 10PCh. 19 - Prob. 11PCh. 19 - Prob. 12PCh. 19 - Prob. 13PCh. 19 - Prob. 14PCh. 19 - Prob. 15PCh. 19 - Prob. 16PCh. 19 - Prob. 17PCh. 19 - Prob. 18PCh. 19 - Prob. 19PCh. 19 - Prob. 20PCh. 19 - Prob. 21PCh. 19 - Prob. 22PCh. 19 - Prob. 23PCh. 19 - Prob. 24PCh. 19 - Prob. 25PCh. 19 - Prob. 26PCh. 19 - Prob. 27PCh. 19 - Prob. 28PCh. 19 - Prob. 29PCh. 19 - Prob. 30PCh. 19 - Prob. 31PCh. 19 - Prob. 32PCh. 19 - Prob. 33PCh. 19 - Prob. 34PCh. 19 - Prob. 35PCh. 19 - Prob. 36PCh. 19 - Prob. 37PCh. 19 - Prob. 38PCh. 19 - Prob. 39PCh. 19 - Prob. 40PCh. 19 - Prob. 41PCh. 19 - Prob. 42PCh. 19 - Prob. 43PCh. 19 - Prob. 44PCh. 19 - Prob. 45PCh. 19 - Prob. 46PCh. 19 - Prob. 47PCh. 19 - Prob. 48PCh. 19 - Prob. 49PCh. 19 - Prob. 50PCh. 19 - Prob. 51PCh. 19 - Prob. 52PCh. 19 - Prob. 53PCh. 19 - Prob. 54PCh. 19 - Prob. 55PCh. 19 - Prob. 56PCh. 19 - Prob. 57PCh. 19 - Prob. 58PCh. 19 - Prob. 59PCh. 19 - Prob. 60PCh. 19 - Prob. 61PCh. 19 - Prob. 62PCh. 19 - Prob. 63PCh. 19 - Prob. 64GPCh. 19 - Prob. 65GPCh. 19 - Prob. 66GPCh. 19 - Prob. 67GPCh. 19 - Prob. 68GPCh. 19 - Prob. 69GPCh. 19 - Prob. 70GPCh. 19 - Prob. 71GPCh. 19 - Prob. 72GPCh. 19 - Prob. 73GPCh. 19 - Prob. 74GPCh. 19 - Prob. 75GPCh. 19 - Prob. 76GPCh. 19 - Prob. 77GPCh. 19 - Prob. 78GPCh. 19 - Prob. 79GPCh. 19 - Prob. 80GPCh. 19 - Prob. 81GPCh. 19 - Prob. 82GPCh. 19 - Prob. 83GPCh. 19 - Prob. 84GPCh. 19 - Prob. 85GPCh. 19 - Prob. 86GPCh. 19 - Prob. 87GP
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