Problem 14
Question
When Kirchhoff's laws are applied to the electrical circuit shown in Figure \(9.1\) the currents \(I_{1}\) and \(I_{2}\) are connected by the equations: $$ \begin{aligned} 27 &=1.5 I_{1}+8\left(I_{1}-I_{2}\right) \\ -26 &=2 I_{2}-8\left(I_{1}-I_{2}\right) \end{aligned} $$ Solve the equations to find the values of currents \(I_{1}\) and \(I_{2}\)
Step-by-Step Solution
Verified Answer
The currents are \(I_1 = 2\) and \(I_2 = -1\).
1Step 1: Simplify the Equations
First, let's expand the terms in each equation:For the first equation: \[ 27 = 1.5I_1 + 8(I_1 - I_2) \]Expand to get:\[ 27 = 1.5I_1 + 8I_1 - 8I_2 \]This simplifies to:\[ 27 = 9.5I_1 - 8I_2 \]For the second equation:\[ -26 = 2I_2 - 8(I_1 - I_2) \]Expand to get:\[ -26 = 2I_2 - 8I_1 + 8I_2 \]This simplifies to:\[ -26 = -8I_1 + 10I_2 \]
2Step 2: Rearrange the Equations
Now, let's rearrange both equations so that we can use them for elimination or substitution:Equation 1: \[ 9.5I_1 - 8I_2 = 27 \]Equation 2: \[ -8I_1 + 10I_2 = -26 \]
3Step 3: Eliminate One Variable
We will multiply Equation 2 by \(1.1875\) to align the coefficients of \(I_1\) in both equations and eliminate one variable:Multiply Equation 2 by \(1.1875\):\[ -9.5I_1 + 11.875I_2 = -30.875 \]Now, add Equation 1 and the new Equation 2:\[ 9.5I_1 - 8I_2 - 9.5I_1 + 11.875I_2 = 27 - 30.875 \]This simplifies to:\[ 3.875I_2 = -3.875 \]
4Step 4: Solve for Remaining Variable
Solve for \(I_2\):\[ I_2 = \frac{-3.875}{3.875} = -1 \]
5Step 5: Substitute Back to Find Other Variable
Substitute \(I_2 = -1\) back into Equation 1:\[ 9.5I_1 - 8(-1) = 27 \]This becomes:\[ 9.5I_1 + 8 = 27 \]Simplify to find:\[ 9.5I_1 = 19\]Solve for \(I_1\):\[ I_1 = \frac{19}{9.5} = 2 \]
6Step 6: Verification Step
It's always good practice to verify the solution by plugging values back into the original equations:First equation:\[ 27 = 1.5 \times 2 + 8(2 - (-1)) = 3 + 8 \times 3 = 27 \]It checks out.Second equation:\[ -26 = 2 \times (-1) - 8(2 - (-1)) = -2 - 8 \times 3 = -26 \]It checks out.
Key Concepts
Electrical Circuit AnalysisSystem of Linear EquationsCurrent CalculationStep-by-Step Problem Solving
Electrical Circuit Analysis
When dealing with electrical circuits, understanding the flow of current is crucial. Kirchhoff's laws provide a systematic way to analyze the circuits and determine the unknown currents and voltages. This involves applying two main principles:
- Kirchhoff's Current Law (KCL): This law states that the sum of currents entering a junction must equal the sum of currents leaving it. In other words, all current entering and leaving a node must add up to zero.
- Kirchhoff's Voltage Law (KVL): This law states that the sum of the electrical potential differences (voltage) around any closed loop in a circuit should be zero.
System of Linear Equations
In many electrical circuit problems, including the one described above, you often end up with a system of linear equations. A system of linear equations is a set of equations with multiple variables that have a linear relationship.
The goal is to find the values of those variables that satisfy all given equations. Each equation represents a line, and solving the system means finding the point(s) at which these lines intersect, which corresponds to the current values in our circuit analysis.
The goal is to find the values of those variables that satisfy all given equations. Each equation represents a line, and solving the system means finding the point(s) at which these lines intersect, which corresponds to the current values in our circuit analysis.
- Start by expressing each equation in standard form.
- Rearrange terms so that similar variables line up across equations.
- Decide on a method to solve them: substitution, elimination, or matrix methods like Gaussian elimination.
Current Calculation
Once an equations system is established through electrical circuit analysis, the next step is calculating the unknown currents. Let's take a closer look at how this was done using the example from the problem:
Begin by simplifying and rearranging your equations for easier calculation. During this process, you can expand terms, combine like terms, and eventually isolate the variables of interest.
In our example:
Begin by simplifying and rearranging your equations for easier calculation. During this process, you can expand terms, combine like terms, and eventually isolate the variables of interest.
In our example:
- Equations were simplified and written in a form that's ready for solving using either substitution or elimination methods.
- We adjusted and lined up terms for both equations to systematically eliminate one variable.
- By multiplying and adding the equations, we isolated a single variable, making it much easier to solve for the remaining unknowns.
Step-by-Step Problem Solving
Solving electrical circuit problems effectively requires a methodical, step-by-step approach. This ensures that all variables are correctly calculated, and helps prevent errors that can occur if steps are skipped.
Here's how you can approach a circuit problem systematically:
Here's how you can approach a circuit problem systematically:
- Step 1: Clearly define all given variables and what you need to find. This might include currents, resistances, and voltages for given elements in the circuit.
- Step 2: Apply Kirchhoff's laws so you can form a system of equations that governs the circuit. Double-check these are accurately representing the circuit's behavior.
- Step 3: Use algebraic methods to simplify and solve these equations, whether by substitution or elimination. Always align the variables across equations to streamline this process.
- Step 4: Verify your solution by substituting calculated values back into the original equations to ensure all conditions are satisfied.
Other exercises in this chapter
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