I Do Not Understand a Textbook Example: Calculating Voltage Across a Resistor

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Solution 1

The book is indeed wrong and the quoted answer $V=0.5\ \text{volts}$ does not appear to make any sense.

Using conservation of energy, one can compare the power going into the source and the power going into the resistor. Obviously for the source this is $$P=VI=(0.002\,\text{A})\cdot V$$ And that for the resistor is $$P=\frac{V^2}{R}=\frac{V^2}{10^3\,\Omega}=(0.001\,\Omega^{-1})\cdot V^2$$ since $$V=IR$$ so $$I=\frac{V}{R}$$

By conservation of energy, these two expressions must be equal meaning $$(0.002\,\text{A})\cdot V=(0.001\,\Omega^{-1})\cdot V^2$$ meaning $$V=2\,\text{A}\,\Omega = 2\ \text{volts}$$

If you calculated the voltage from the resistance and current using Ohm's law then you would get $$V=I\ R=2\times 10^{-3}\,\text{A}\times 1000\,\Omega=2\ \text{volts}$$ which is also consistent with the conservation of energy analysis above.

Solution 2

You're correct. Book is wrong. Ohm's Law and all three power conservation methods give 2V since they are mathematically equivalent:

$$P = \frac{V^2}{R} = I^2{R}$$

$$V^2= I^2{R}^2$$

$$V= IR$$

Or

$$P = VI= I^2{R}$$ $$V= IR$$

Or $$P = \frac{V^2}{R} = VI$$ $$V^2 = VIR$$ $$V= IR$$

Solution 3

Since there's only one loop, the current in the loop is determined by the current source and is equal to $i= 2\,\mathrm{mA}.$ Flowing through the resistance $R=1\,\mathrm{k\Omega},$ it creates a voltage drop $v=i‎‎R=(2/1000)‎‎(1\times1000)=2\,\mathrm{V}$. The book probably has a typo.‎

*This answer was translated from Russian.

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Updated on August 01, 2022

Comments

  • clo_jur
    clo_jur over 1 year

    I am reading Agarwal and Lang's Foundations of Analog and Digital Circuits for self study, and I do not understand one of their illustrated examples (specifically, example 2.14).

    They present this circuit and claim $v = 0.5$ V. If I were to use a different method, namely the fact that $v = iR$, why am I incorrect that $v = 2$ mA $\times$ $1$ k$\Omega = 2$ V?

    The book uses an "energy conservation" method which yields $v = 0.5$ V.

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    • rob
      rob about 2 years
      A number of comments removed. Friendly reminder that comments are for improving the question; to answer the question, post an answer.
    • Scott Seidman
      Scott Seidman about 2 years
      If you post a synopsis of the energy conservation solution offered, perhaps we can highlight the exact mistake in the solution.
    • muru
      muru about 2 years
      @ScottSeidman I had a look at what seems to be a PDF of this book I found when searching its name, and the concluding lines of this example are: "Finally, equating the power supplied by the source to the power dissipated by the resistor, we have $$0.002v = 0.001v^2.$$ In other words, $v = 0.5 V$." It's clearly an accident.
    • PcMan
      PcMan about 2 years
      It is sad when a $100 educational book issued by two professors at MIT, contains such ludicrous errors. Here the error is obvious and fixable, but how many other untruths are hidden in its pages?
  • Buzz
    Buzz about 2 years
    Comments are not for extended discussion; this conversation has been moved to chat.