Find I And Vo In The Circuit Of Fig 2.100

Let's delve into circuit analysis to determine the values of current i and voltage v₀ in the circuit depicted in Fig 2.100, a fundamental task in electrical engineering.

Circuit Analysis: Finding i and v₀ in Fig 2.100

Solving for unknown currents and voltages in a circuit is a core skill for electrical engineers and technicians. This process relies on applying fundamental circuit laws and theorems to simplify the circuit and solve for the desired variables. The specific approach depends on the complexity of the circuit, but common techniques include Ohm's Law, Kirchhoff's Laws (KVL and KCL), series and parallel resistance combinations, and potentially more advanced methods like nodal or mesh analysis. Understanding these principles is crucial for designing, analyzing, and troubleshooting electrical systems.

Let’s assume Fig 2.100 represents a circuit containing resistors and voltage/current sources. Without the actual diagram, we'll create a general problem that highlights key concepts. Assume the circuit consists of the following components:

A voltage source, V = 24V.
Three resistors: R1 = 4Ω, R2 = 8Ω, and R3 = 12Ω.
The resistors are connected in a series-parallel combination. R2 and R3 are in parallel, and this parallel combination is in series with R1 and the voltage source.

We want to find:

i: The current flowing through the voltage source and R1.
v₀: The voltage across R3.

Step-by-Step Solution

Simplify the Circuit:

The first step is to simplify the circuit by finding the equivalent resistance of the parallel combination of R2 and R3. The formula for the equivalent resistance (Req) of two resistors in parallel is:

1/Req = 1/R2 + 1/R3

Plugging in the values:

1/Req = 1/8 + 1/12 = (3 + 2) / 24 = 5/24

Therefore, Req = 24/5 = 4.8Ω.
Calculate the Total Resistance:

Now that we have the equivalent resistance of the parallel combination, we can find the total resistance (Rtotal) of the entire circuit. Since Req is in series with R1, we simply add them:

Rtotal = R1 + Req = 4Ω + 4.8Ω = 8.8Ω.
Calculate the Current (i):

Using Ohm's Law (V = IR), we can find the current i flowing through the circuit:

i = V / Rtotal = 24V / 8.8Ω ≈ 2.73A.
Calculate the Voltage Across the Parallel Combination:

The voltage across the parallel combination (which is also the voltage across R3, i.e., v₀) can be found using Ohm's Law again, but this time using the equivalent resistance and the current i:

v₀ = i * Req = 2.73A * 4.8Ω ≈ 13.10V.

Therefore, v₀ ≈ 13.10V.

Detailed Explanation of Each Step

Step 1: Simplifying the Circuit (Parallel Resistors)

The resistors R2 and R3 are connected in parallel. This means that the voltage across both resistors is the same. The combined effect of these parallel resistors is to provide a lower overall resistance than either resistor alone. The formula 1/Req = 1/R2 + 1/R3 arises from the fact that in a parallel circuit, the total current divides between the branches, with each branch carrying a current inversely proportional to its resistance. The reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances. This simplification allows us to treat the parallel combination as a single equivalent resistor, making the overall circuit analysis easier. Accurately calculating the equivalent resistance is crucial, as it directly affects the calculation of the total current.

Step 2: Calculating the Total Resistance (Series Resistors)

Once the parallel combination is simplified to an equivalent resistor (Req), this equivalent resistance is in series with resistor R1. In a series circuit, the current is the same through all components. The total resistance of series resistors is simply the sum of their individual resistances. This is because the current must flow through each resistor sequentially, encountering the resistance of each one along the way. Adding R1 and Req gives us the total resistance that the voltage source "sees," which is essential for determining the total current supplied by the source. Failing to correctly calculate the total resistance will lead to errors in subsequent current and voltage calculations.

Step 3: Calculating the Current (i) (Ohm's Law)

Ohm's Law (V = IR) is a fundamental relationship in circuit theory, stating that the voltage across a resistor is directly proportional to the current flowing through it. The constant of proportionality is the resistance. By rearranging the formula to i = V/R, we can calculate the current if we know the voltage and resistance. In this case, we use the total voltage supplied by the source (V) and the total resistance of the circuit (Rtotal) to find the total current (i) flowing from the source. This current is the same current that flows through R1 because it is in series with the voltage source. This step is a direct application of Ohm's Law to the entire circuit, and it is the key to finding the initial current value needed for further analysis.

Step 4: Calculating the Voltage Across R3 (v₀) (Ohm's Law)

The voltage across the parallel combination of R2 and R3 (which is v₀, the voltage across R3) is equal to the current flowing through the equivalent resistance of the parallel combination (Req) multiplied by Req. This is another application of Ohm's Law (V = IR), but this time applied to a specific portion of the circuit. Since R2 and R3 are in parallel, they have the same voltage across them. Therefore, calculating the voltage across Req directly gives us the voltage across R3. This step demonstrates how to apply Ohm's Law to a simplified portion of the circuit to find specific voltage values.

Alternative Methods and Considerations

While the above method is straightforward, other techniques can be used depending on the complexity of the original Fig 2.100:

Nodal Analysis: This method involves identifying nodes (points where multiple circuit elements connect) and writing Kirchhoff's Current Law (KCL) equations at each node. KCL states that the sum of currents entering a node must equal the sum of currents leaving the node. Solving the system of equations yields the node voltages, from which currents and other voltages can be determined. Nodal analysis is particularly useful for circuits with multiple voltage sources.
Mesh Analysis: This method involves identifying loops (closed paths) in the circuit and writing Kirchhoff's Voltage Law (KVL) equations for each loop. KVL states that the sum of the voltages around a closed loop must equal zero. Solving the system of equations yields the loop currents, which can then be used to find branch currents and voltages. Mesh analysis is particularly useful for circuits with multiple current sources.
Superposition Theorem: This theorem states that in a linear circuit with multiple independent sources, the response (current or voltage) at any point in the circuit is the sum of the responses due to each independent source acting alone, with all other independent sources turned off (voltage sources replaced with short circuits, and current sources replaced with open circuits). This can be useful for simplifying the analysis of circuits with multiple sources.
Source Transformation: This technique allows you to convert a voltage source in series with a resistor into a current source in parallel with the same resistor, or vice versa. This can sometimes simplify the circuit and make it easier to analyze using other methods.

Importance of Understanding Circuit Analysis

Understanding circuit analysis techniques is essential for:

Designing Electrical Circuits: Engineers use circuit analysis to design circuits that meet specific performance requirements. This includes determining the appropriate values for components, such as resistors, capacitors, and inductors, to achieve desired voltage and current levels.
Troubleshooting Electrical Systems: When electrical systems malfunction, circuit analysis can be used to identify the cause of the problem. By measuring voltages and currents at different points in the circuit, technicians can pinpoint the faulty component.
Analyzing Electronic Devices: Circuit analysis is also used to analyze the behavior of electronic devices, such as amplifiers, filters, and oscillators. This helps engineers understand how these devices work and how to improve their performance.
Power System Analysis: Analyzing complex power grids relies heavily on network analysis techniques derived from basic circuit analysis principles.

Practical Applications

The principles illustrated in this example problem are applicable in a wide range of real-world scenarios:

Lighting Circuits: Determining the correct resistor values to limit current to LEDs in a lighting circuit.
Audio Amplifiers: Calculating voltage and current gains in amplifier stages.
Power Supplies: Designing voltage dividers to create specific voltage levels for different components in a power supply.
Sensor Circuits: Analyzing the output of sensor circuits and calibrating them for accurate measurements.
Control Systems: Modeling and analyzing control systems that use electrical circuits to regulate physical processes.

Common Mistakes to Avoid

Incorrectly Applying Ohm's Law: Ensure you are using the correct voltage, current, and resistance values for the component or portion of the circuit you are analyzing.
Misidentifying Series and Parallel Connections: Accurately identifying series and parallel connections is crucial for simplifying the circuit correctly.
Forgetting to Account for Voltage or Current Dividers: In series circuits, voltage divides proportionally across resistors. In parallel circuits, current divides inversely proportionally to resistors.
Using the Wrong Units: Always use consistent units (volts, amps, ohms) when performing calculations.
Ignoring Polarity: Pay attention to the polarity of voltage sources and the direction of current flow.

Advanced Techniques (Beyond the Scope of this Example)

For more complex circuits, you might need to employ more advanced techniques:

Thevenin's Theorem: This theorem allows you to replace a complex circuit with a simpler equivalent circuit consisting of a voltage source and a series resistor.
Norton's Theorem: This theorem allows you to replace a complex circuit with a simpler equivalent circuit consisting of a current source and a parallel resistor.
Laplace Transforms: These transforms are used to analyze circuits with time-varying signals.
Phasor Analysis: This technique is used to analyze AC circuits in the frequency domain.

Example with Dependent Sources

Let's consider a slightly more complex scenario. Suppose Fig 2.100 contains a dependent voltage source. A dependent source's voltage or current depends on a voltage or current elsewhere in the circuit.

Assume:

V = 24V (Independent Voltage Source)
R1 = 4Ω
R2 = 8Ω
R3 = 12Ω
A dependent voltage source: 2 * iₓ, where iₓ is the current flowing through R1. This makes it a current-controlled voltage source (CCVS). The dependent source is in series with R3.

We still want to find i (current through V and R1) and v₀ (voltage across R3).

Modified Solution:

Define Variables:
- i = current through R1 and the independent voltage source V.
- i₂ = current through R2.
- i₃ = current through R3 and the dependent voltage source.
- v₀ = voltage across R3.
- Since i = iₓ, the dependent source is 2i.
Apply Kirchhoff's Laws:
- KVL Loop 1 (V - R1 - R2): V - i * R1 - i₂ * R2 = 0 => 24 - 4i - 8i₂ = 0
- KVL Loop 2 (R2 - R3 - Dependent Source): i₂ * R2 - i₃ * R3 - 2i = 0 => 8i₂ - 12i₃ - 2i = 0
- KCL at the Node Connecting R1, R2, and R3: i = i₂ + i₃
Solve the System of Equations:

We now have a system of three equations with three unknowns (i, i₂, i₃):
- Equation 1: 4i + 8i₂ = 24 => i + 2i₂ = 6
- Equation 2: 2i + 12i₃ - 8i₂ = 0
- Equation 3: i = i₂ + i₃
Solve for i, i₂, and i₃. This can be done using substitution or matrix methods.

From Equation 1: i = 6 - 2i₂

Substitute into Equation 2: 2(6 - 2i₂) + 12i₃ - 8i₂ = 0 => 12 - 4i₂ + 12i₃ - 8i₂ = 0 => 12i₃ - 12i₂ = -12 => i₃ - i₂ = -1

Substitute into Equation 3: 6 - 2i₂ = i₂ + i₃ => 6 = 3i₂ + i₃

Now we have two equations:
- i₃ - i₂ = -1
- 3i₂ + i₃ = 6
Subtract the first equation from the second: 4i₂ = 7 => i₂ = 7/4 = 1.75 A

Substitute i₂ back into i₃ - i₂ = -1: i₃ - 1.75 = -1 => i₃ = 0.75 A

Substitute i₂ back into i = 6 - 2i₂: i = 6 - 2(1.75) = 6 - 3.5 = 2.5 A

Therefore, i = 2.5 A
Calculate v₀:

v₀ = i₃ * R3 = 0.75 A * 12Ω = 9V

Therefore, v₀ = 9V.

Key Difference with Dependent Sources:

The presence of a dependent source significantly complicates the analysis. You can no longer directly combine resistors as easily because the voltage or current relationships are tied to other parts of the circuit. You must use KVL and KCL to create a system of equations and solve for the unknowns. The dependent source introduces a coupling between different parts of the circuit, making the problem more mathematically involved.

Conclusion

Finding the current i and voltage v₀ in a circuit, like the one represented by Fig 2.100 (or our generalized example), involves applying fundamental circuit analysis techniques. This includes simplifying the circuit using series and parallel resistance combinations, applying Ohm's Law, and, for more complex circuits, using Kirchhoff's Laws (KVL and KCL). The presence of dependent sources requires a more rigorous approach using KVL and KCL to solve a system of equations. A solid understanding of these principles is crucial for anyone working with electrical circuits, enabling them to design, analyze, and troubleshoot a wide variety of electrical systems. Consistent practice and a methodical approach are key to mastering these skills.