Help me to understand basic circuit analysis solution

[General_Sax]

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!

 

[skeptic2]

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?

 

[General_Sax]

That makes much more sense. Thanks for your time!

 

[evra]

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

 

[DeeNos]

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