Always double-check the visual polarities ($+$ and $-$ placement) of the voltage sources in schematic diagrams. Here, the middle source has its positive terminal connected to the ground rail and its negative terminal facing upwards, making its node potential contribution negative!
Concept:
We can calculate the voltage at node P, denoted as $V_P$, by setting the bottom reference rail wire as ground ($0\text{V}$) and applying Kirchhoff's Current Law (KCL) at node P. KCL dictates that the total algebraic sum of all currents leaving a node must equal zero.
Step-by-step Nodal Analysis Formulation:
Let us analyze each individual vertical branch connected to node P:
• Left Branch (Branch 1): Contains a $2\Omega$ resistor ($R_1$) and a $4\text{V}$ independent voltage source ($V_1$). The current leaving node P downward through this branch is:
$$I_{\text{left}} = \frac{V_P - V_1}{R_1} = \frac{V_P - 4}{2}$$
• Middle Branch (Branch 2): Contains a $3\Omega$ resistor ($R_2$) and a $6\text{V}$ independent voltage source ($V_2$). The current leaving node P downward through this branch is:
$$I_{\text{middle}} = \frac{V_P - V_2}{R_2} = \frac{V_P - 6}{3}$$
• Right Branch (Branch 3): Contains a $1\Omega$ resistor ($R_3$) in series with a $3\text{A}$ current source ($I_1$) pointing upwards towards node P. Since the current source forces $3\text{A}$ to enter node P, the current leaving node P through this branch is explicitly:
$$I_{\text{right}} = -3\text{ A}$$
(Note: The value of the $1\Omega$ series resistor does not alter the current, as a current source maintains its specified current regardless of series resistance).
Step 2: Solving the KCL Equation.
Summing these three current expressions according to KCL:
$$\frac{V_P - 4}{2} + \frac{V_P - 6}{3} - 3 = 0$$
To eliminate the denominators, multiply the entire equation by the Least Common Multiple (LCM) of 2 and 3, which is 6:
$$6 \cdot \left( \frac{V_P - 4}{2} \right) + 6 \cdot \left( \frac{V_P - 6}{3} \right) - 6 \cdot (3) = 0$$
$$3(V_P - 4) + 2(V_P - 6) - 18 = 0$$
Expand the brackets:
$$3V_P - 12 + 2V_P - 12 - 18 = 0$$
Group all matching $V_P$ terms and constant numerical terms together:
$$(3 + 2)V_P - (12 + 12 + 18) = 0$$
$$5V_P - 42 = 0$$
$$5V_P = 42$$
$$V_P = \frac{42}{5} = 8.4\text{V}$$
*Correction Note on Option Matching:* Let's look closer at the voltage source polarity signs in the diagram. The 6V source is oriented with its negative terminal upwards, meaning it is $-6\text{V}$ relative to ground. Let's recalculate with the negative terminal up:
$$I_{\text{middle}} = \frac{V_P - (-6)}{3} = \frac{V_P + 6}{3}$$
Let's re-evaluate the equation:
$$\frac{V_P - 4}{2} + \frac{V_P + 6}{3} - 3 = 0$$
Multiply by 6:
$$3(V_P - 4) + 2(V_P + 6) - 18 = 0$$
$$3V_P - 12 + 2V_P + 12 - 18 = 0$$
$$5V_P - 18 = 0$$
$$5V_P = 18$$
$$V_P = \frac{18}{5} = 3.6\text{V}$$
This matches Option (B) perfectly.
Was this answer helpful?
0
0
Top AP PGECET Electronics and Communication Engineering Questions
