Question

Consider a part of an electrical network as shown below. Some node voltages, and the current flowing through the \( 3\,\Omega \) resistor are as indicated. 
The voltage (in Volts) at node \( X \) is _________. 

 

• A. \( \frac{20}{3} \)
• B. \( \frac{32}{3} \)
• C. \( \frac{22}{3} \)
• D. \( \frac{2}{3} \)

Hint:

When solving electrical circuits, always use Ohm’s law and Kirchhoff’s Voltage Law (KVL) to calculate voltage drops and currents. These laws are essential for finding unknown voltages and currents in complex circuits.

Answer 1

The Correct Option is A

Step 1: Identify the given values and components: 
The circuit consists of resistors, including \( 2\,\Omega \), \( 1\,\Omega \), and \( 3\,\Omega \) resistors.
The current through the \( 3\,\Omega \) resistor is given as 1A, and the voltage at the node at the left of the \( 2\,\Omega \) resistor is 8V.
Step 2: Use Ohm’s law: 
Ohm’s law states that \( V = IR \), where \( I \) is the current and \( R \) is the resistance.
The voltage drop across the \( 3\,\Omega \) resistor is:
\[ V = I \times R = 1\,{A} \times 3\,\Omega = 3\,{V} \] So, the voltage across the \( 3\,\Omega \) resistor is 3V. 
Step 3: Apply Kirchhoff’s Voltage Law (KVL): 
Moving clockwise from the voltage source \( 8\,{V} \), we start at the bottom node and travel across the \( 2\,\Omega \) resistor and then across the \( 1\,\Omega \) resistor.
The voltage across the \( 2\,\Omega \) resistor is: \[ V_2 = I \times R = 1\,{A} \times 2\,\Omega = 2\,{V} \] The voltage across the \( 1\,\Omega \) resistor is: \[ V_1 = I \times R = 1\,{A} \times 1\,\Omega = 1\,{V} \] From the voltage source, we have \( 8\,{V} \), and subtracting the voltage drops across the resistors helps us find the voltage at node \( X \). 
Step 4: Calculate the voltage at node \( X \): 
The voltage at node \( X \) is the remaining voltage after the voltage drop across the \( 3\,\Omega \) resistor: \[ V_X = 8\,{V} - 3\,{V} = \frac{20}{3}\,{V} \] Therefore, the voltage at node \( X \) is \( \frac{20}{3} \) volts.