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I have the following circuit, and have asked to be found the potential Vo:

schematic

simulate this circuit – Schematic created using CircuitLab

Using nodal analysis and Kirchoff's Current Law, we have:

At Node V0: $$\frac{V_0}{R_5}+\frac{V_0 - V_1}{R_4}=2\times 10^{-3} \\ \therefore \:3V_0 - V_1 = 4$$

At Node V1: $$\frac{V_1 - V_0}{R_4}+\frac{V_1}{R_3}+\frac{V_1-V_2}{R_1}=0 \\ \therefore \: -3V_0 + 5V_1 - V_2 = 0$$

At Node V2: $$\frac{V_2 - V_1}{R_1} + \frac{V_2}{R_2}=6\times 10^{-3} \\ \therefore \: -2V_1 + 3V_2 = 36$$

Hence we have the matrix equation:

$$\begin{bmatrix}3 & -1 & 0 \\ -3 & 5 & -1 \\ 0 & -2 & 3\end{bmatrix}\begin{bmatrix}V_0 \\ V_1 \\ V_2\end{bmatrix} = \begin{bmatrix}4 \\ 0 \\ 36\end{bmatrix}$$

Therefore we can solve to yield:

$$V_0 = \frac{44}{15} \approx 2.9333, V_1 = \frac{24}{5}, V_2 = \frac{76}{5}$$

However, the book that I am working from (Basic Engineering Circuit Analysis, Tenth Edition; Irwin and Nelms) states that: $$V_0 = 2.79 \mathrm{V}$$

What have I done wrong here or is this a typo in the book (I am unsure because there have been previous typos such as 60 seconds in an hour)?

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  • \$\begingroup\$ What does the CircuitLab simulation say? \$\endgroup\$ – Joe Hass Dec 14 '13 at 12:37
  • \$\begingroup\$ @JoeHass I just ran it and it says that $$V_0 = 2.79 \mathrm{V}$$ Which at least means the book is correct; however, I still am not sure where my error is! \$\endgroup\$ – Thomas Russell Dec 14 '13 at 12:45
  • \$\begingroup\$ Check your algebra for the simplification of the V2 KCL equation. \$\endgroup\$ – Alfred Centauri Dec 14 '13 at 13:47
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Make sure you are consistent when defining current though the resistor R4: at node V0 you have it going towards V1 at V1 you have it going towards V0.

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  • \$\begingroup\$ Forgive me for my ignorance (I'm really quite new at Circuit analysis), but I was under the impression that consistency was only required at each individual node? The choice of direction of current at different nodes is independent? \$\endgroup\$ – Thomas Russell Dec 14 '13 at 16:41
  • \$\begingroup\$ No, you have to have consistency throughout: y=2x-5 5x=1 is a different system than -y=2x-5 5x=1 So, I would label every current as going in a particular direction and then stick to that convention \$\endgroup\$ – Yuriy Dec 14 '13 at 18:38
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    \$\begingroup\$ @Shaktal, Yuriy is incorrect; your KCL equations are correctly done but, you made a simple math error when simplifying the V2 KCL equation. \$\endgroup\$ – Alfred Centauri Dec 14 '13 at 20:22

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