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So I'm trying to solve a bunch of exercises but I always end up with the opposite sign for the voltage than the correct answer.

The question is to find the voltages of the nodes.

Circuit diagram drawn in LTSpice

I'm trying to solve it using KCL for each node like this.

\$v_1\$ is the leftmost node above \$R1\$.
\$v_2\$ is the middle node above \$R2\$.
\$v_3\$ is the rightmost node above \$R3\$.

Equation 1: KCL for \$v_1\$
\$\frac{v_1}{5\Omega} + \frac{v_1 - v_2}{4\Omega} - 3.5A = 0 \Rightarrow v_1 = \frac{70}{9} V + \frac{5}{9} v_2\$

Equation 2: KCL for \$v_2\$
\$3.5A + \frac{v_2 - v_1}{4\Omega} + \frac{v_2}{2.5\Omega} + \frac{v_2 - v_3}{5\Omega} = 0 \Rightarrow 3.5A + \frac{17}{20\Omega} v_2 - \frac{1}{4\Omega} v_1 - \frac{1}{5\Omega} v_3 = 0\$

Equation 3: KCL for \$v_3\$
\$2A + \frac{v_3}{10\Omega} + \frac{v_3 - v_2}{5\Omega} = 0 \Rightarrow - \frac{20}{3} V + \frac{2}{3} v_2 = v_3\$

Solving for \$v_2\$ by plugging in equation 1 and 3 into equation 2 like this
\$3.5A + \frac{17}{20\Omega} v_2 - \frac{1}{4\Omega} (\frac{70}{9} V + \frac{5}{9} v_2) - \frac{1}{5\Omega} (- \frac{20}{3} V + \frac{2}{3} v_2) = 0 \Rightarrow v_2 = -5V\$

The answer according to the circuit simulator is \$5V\$. The book confirms that \$5V\$ is the correct answer for \$v_2\$.

What am I doing wrong?

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    \$\begingroup\$ Sum of currents away from a node = 0; e.g. Eqn 1 should be +3.5 \$\endgroup\$ – Chu May 24 '17 at 20:23
  • \$\begingroup\$ You shouldn't put units in equations - it's really confusing. So omit the \$\small A\$, \$\small V\$, \$\small \Omega\$ etc. \$\endgroup\$ – Chu May 25 '17 at 6:48
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Chu's comment is the right tip. To answer what you're doing wrong in a more general sense: You only have the direction of the current sources reversed compared to your calcuated (Voltage/Resistance) currents.

Starting with Eqn 1, v1/5 is current flowing from v1 to ground. The 3.5A current source is also moving away from v1. So, if you're going to list all the current sources coming from v1, they should be the same sign (+) and have them added together (v1/5 + 3.5 + (v1-v2)/4).

Similarly for Eqn 2 & 3, you just have the sign/polarity of the current sources inconsistent with your V/R currents. So, -3.5A in Eqn2, and -2A in Eqn3.

Your consistent reversal of the current sources would explain your consistency in getting negative versions of the right answer.

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I prefer to list out-going currents on the left and incoming currents on the right. It helps me mentally "book-keep" more easily:

$$\begin{align*} \frac{V_1}{R_1}+\frac{V_1}{R_5}+3.5\:\textrm{A}&=\frac{0\:\textrm{V}}{R_1}+\frac{V_2}{R_5}\\\\ \therefore 0 &=\frac{V_1}{R_1}+\frac{V_1-V_2}{R_5}+3.5\:\textrm{A}\\\\ \frac{V_2}{R_2}+\frac{V_2}{R_4}+\frac{V_2}{R_5}&=3.5\:\textrm{A}+\frac{0\:\textrm{V}}{R_2}+\frac{V_3}{R_4}+\frac{V_1}{R_5}\\\\ \therefore 0 &=\frac{V_2}{R_2}+\frac{V_2-V_3}{R_4}+\frac{V_2-V_1}{R_5}-3.5\:\textrm{A}\\\\ \frac{V_3}{R_3}+\frac{V_3}{R_4}&=2\:\textrm{A}+\frac{0\:\textrm{V}}{R_3}+\frac{V_2}{R_4}\\\\ \therefore 0 &=\frac{V_3}{R_3}+\frac{V_3-V_2}{R_4}-2\:\textrm{A} \end{align*}$$

From the above, you can see several errors you made related to the current sources.

This solves out as: \$V_1=-5\:\textrm{V}\$, \$V_2=5\:\textrm{V}\$, \$V_3=10\:\textrm{V}\$.

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