Chapter 3, Problem 1RQ

At node 1 in the circuit of Fig. 3.46, applying KCL gives:

(a)     $2+\frac{12-v_1}{3}=\frac{v_1}{6}+\frac{v_1-v_2}{4}$

(b)     $2+\frac{v_1-12}{3}=\frac{v_1}{6}+\frac{v_2-v_1}{4}$

(c)     $2+\frac{12-v_1}{3}=\frac{0-v_1}{6}+\frac{v_1-v_2}{4}$

(d)     $2+\frac{v_1-12}{3}=\frac{0-v_1}{6}+\frac{v_2-v_1}{4}$

To determine

Choose the correct option to find the node equation at node 1 in the circuit of Figure 3.46.

Answer to Problem 1RQ

The correct option from the given choices is option $\text{(a)}2+\frac{12-v_1}{3}=\frac{v_1}{6}+\frac{v_1-v_2}{4}$.

Explanation of Solution

Given data:

Refer Figure 3.46 in the textbook for the nodal analysis.

Formula used:

Write the basic principle for the current flows through resistor.

$i=\frac{v_{\text{higher}}-v_{\text{lower}}}{R}$

Here,

$v_{\text{higher}}$ is the higher potential,

$v_{\text{lower}}$ is the lower potential, and

$R$ is the resistance of the resistor.

Calculation:

Apply Kirchhoff’s current law at node 1 in Figure 3.46.

$\begin{array}{l}2+\frac{12-v_1}{3}-\frac{v_1}{6}-\frac{v_1-v_2}{4}=0\\ 2+\frac{12-v_1}{3}=\frac{v_1}{6}+\frac{v_1-v_2}{4}\end{array}$

Hence, the node equation at node 1 is $2+\frac{12-v_1}{3}=\frac{v_1}{6}+\frac{v_1-v_2}{4}$. Therefore, the option (a) is correct and the options (b), (c), (d) are incorrect.

Conclusion:

Thus, the correct option from the given choices is option $\text{(a)}2+\frac{12-v_1}{3}=\frac{v_1}{6}+\frac{v_1-v_2}{4}$.