How Do You Solve Complex Circuit Problems?

[nn3568]

Homework Statement

Consider the circuit in the figure.

Find the current in the 1.3 Ω resistor. Answer in units of A.
Find the potential difference across the 1.3 Ω resistor. Answer in units of V.
Find the current in the 10.0 Ω resistor. Answer in units of A.

Homework Equations

V = IR
R_eq = 1/R₁ + 1/R₂ + 1/R₃ ...

The Attempt at a Solution

First, I calculated the equivalent resistance.

The top path - parallel circuit: 1/((1/7.7) + (1/7.7)) = 3.85 + 3.4 = 7.25​

The middle path - parallel circuit: 1/((1/4.6) + (1/10.0)) = 3.150684932 + 1.3 = 4.450684932​

R_eq = ?​

My teacher helped me at school and when I tried to do the problem on my own, using his steps, I forgot them all. Just yesterday I remember that I could get the equivalent resistance which is around 6.157 but I keep trying and trying but I can't seem to get that answer either. I need help on that part too.

Here is the total current: I = V / R -> I = 18.0 / 6.157 = approx. 2.92

I don't really understand how to find the current etc. It gives me headaches! Sorry the picture is kinda dark...

 

[Carid]

Everything you have done is OK. Except the equation is

1/Req = 1/R1 + 1/R2 +...

Treat the top and middle paths as two parallel resistances and find their equivalent resistance before adding the final 3.4 Ohm. You will get the 6.157 Ohm value.

Then to answer questions 1 and 2 you have to work out how much of that 2.92 amperes of current will go though the middle branch.

For question 3 you conduct a similar investigation.

 

[nlightner]

I understand that complex circuit problems can be challenging and require a good understanding of circuit principles and equations. Let's break down the steps to finding the current and potential difference in this circuit.

First, we need to calculate the equivalent resistance for the entire circuit. This can be done by combining the resistors in series and parallel. For the top path, we have two resistors in parallel (7.7 Ω and 7.7 Ω) so the equivalent resistance is 3.85 Ω. For the middle path, we have a resistor in series with a parallel combination (4.6 Ω and 10.0 Ω), so the equivalent resistance is 4.6 Ω. Finally, we have the bottom path with a single 1.3 Ω resistor. The total equivalent resistance for the circuit is then 3.85 Ω + 4.6 Ω + 1.3 Ω = 9.75 Ω.

Next, we can use Ohm's Law (V = IR) to find the current in the 1.3 Ω resistor. Since we know the potential difference (V) is 18.0 V and the equivalent resistance (R) is 9.75 Ω, we can rearrange the equation to solve for the current (I). So, I = V/R = 18.0 V / 9.75 Ω = 1.85 A.

To find the potential difference across the 1.3 Ω resistor, we can use Ohm's Law again. Since we know the current (I) is 1.85 A and the resistance (R) is 1.3 Ω, we can solve for the potential difference (V). So, V = IR = 1.85 A x 1.3 Ω = 2.405 V.

Finally, to find the current in the 10.0 Ω resistor, we can use Ohm's Law once more. Since we know the potential difference (V) is 18.0 V and the resistance (R) is 10.0 Ω, we can solve for the current (I). So, I = V/R = 18.0 V / 10.0 Ω = 1.8 A.

I hope this helps you understand how to approach complex circuit problems. Remember to always start by finding the equivalent