Help me to understand basic circuit analysis solution

In summary, the person is struggling with understanding the solution to a circuit problem involving parallel resistances. They question why (12||6) + 2 is used instead of (12||8) and if there is a specific order to calculating effective resistance. They are given a tip to redraw the circuit and start with the resistances farthest from the source. Another person suggests a simpler method of solving the problem.
  • #1
General_Sax
446
0

Homework Statement


given in uploads.

Homework Equations


The Attempt at a Solution



I just don't understand why they went [(12||6) + 2]

I tried doing (12||8) and I ended up getting the wrong answer.

Why do you have to add the 2k(ohms) to the effective resistance of the 12||6 k(ohms)?

Why couldn't one use 12||8 ?

Another question: Is there any paticular order to calculating the effective resistance?

ie, does one have to work outside -> inside or something similiar?

Please and thank you!
 

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  • #2
Before beginning to work the problem it helps to redraw the circuit to simplify it, at least in your mind, as in the attachment.

It usually is best to start with the resistances farthest from the source and work back towards the source.

So in the attached circuit do you see why first the parallel combination of the 12k and 6k are calculated and then 2k is added?
 

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  • #3
That makes much more sense. Thanks for your time!
 
  • #4
there is a simpler method. i.e;
solve the:
step 1: 2k//6k
step 2: 4k//12
step 3: draw the circuit equivalent whereas u will have the rest in parallel
step 4: u will get exactly 2k
 
  • #5


I understand that learning circuit analysis can be challenging and confusing at times. It is important to remember that in circuit analysis, we are trying to find the effective resistance of a circuit, which is the equivalent resistance that would produce the same current and voltage as the original circuit. In order to do this, we use a combination of series and parallel circuit rules.

In this specific problem, we are given a circuit with a 12k(ohms) resistor in series with a parallel combination of a 6k(ohms) and 2k(ohms) resistor. To find the effective resistance, we first need to simplify the parallel combination into one equivalent resistor. This is where the concept of equivalent resistance in series and parallel circuits comes in.

The equation for equivalent resistance in parallel is 1/Re = 1/R1 + 1/R2, where Re is the equivalent resistance and R1 and R2 are the individual resistances in the parallel combination. In this case, R1 is 6k(ohms) and R2 is 2k(ohms). Plugging these values into the equation, we get 1/Re = 1/6k + 1/2k = 1/3k. Solving for Re, we get Re = 3k(ohms). This is the equivalent resistance of the parallel combination.

Now, we can simplify the circuit by replacing the parallel combination with one 3k(ohms) resistor. This means that the circuit now consists of a series combination of a 12k(ohms) resistor and a 3k(ohms) resistor. To find the effective resistance of this series combination, we simply add the individual resistances, giving us 15k(ohms).

To answer your question about why we add the 2k(ohms) resistor to the effective resistance of the parallel combination, it is because the 2k(ohms) resistor is still part of the overall circuit and contributes to the effective resistance. Using 12||8 would not give us the correct answer because it does not take into account the 2k(ohms) resistor.

As for the order of calculating the effective resistance, there is no specific rule. It is important to first identify any parallel or series combinations and simplify them before finding the overall effective resistance.

I hope this explanation helps you better understand basic circuit analysis. Remember to always
 

FAQ: Help me to understand basic circuit analysis solution

1. What is circuit analysis and why is it important?

Circuit analysis is the process of understanding and solving problems related to electrical circuits. It involves analyzing the behavior of electrical components and determining the flow of current and voltage in a circuit. It is important because it allows us to design and troubleshoot circuits, ensuring that they function properly and safely.

2. What are the basic components of a circuit?

The basic components of a circuit include a power source (such as a battery or power supply), conductors (wires), resistors, capacitors, and inductors. These components work together to provide a path for the flow of electrical current.

3. How do I solve a basic circuit analysis problem?

To solve a basic circuit analysis problem, you will need to use a combination of Ohm's law and Kirchhoff's laws. Ohm's law states that the current through a conductor is directly proportional to the voltage and inversely proportional to the resistance. Kirchhoff's laws state that the sum of currents entering a node is equal to the sum of currents leaving the node, and the sum of voltage drops in a closed loop is equal to the sum of voltage sources in that loop. By applying these laws and using algebraic equations, you can solve for the unknown values in a circuit.

4. What tools do I need for basic circuit analysis?

Some essential tools for basic circuit analysis include a voltmeter, ammeter, and multimeter. These tools are used to measure voltage, current, and resistance, respectively. You will also need a basic understanding of mathematics, including algebra and trigonometry, to solve circuit analysis problems.

5. How can I improve my understanding of basic circuit analysis?

The best way to improve your understanding of basic circuit analysis is by practicing. Start with simple circuits and work your way up to more complex ones. You can also use online resources, textbooks, and tutorials to learn about different techniques and strategies for solving circuit analysis problems. Additionally, joining a study group or seeking help from a mentor or tutor can also be beneficial.

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