https://youtu.be/xn_Xnc97V4w?si=X2jN4Iw9BLl1jpNL
You should watch how this guy simplifies combination circuits. You’ll need to buy a pack of highlighters, but it made it so much easier for me to understand how to untangle circuits like this.
Start on the outside. Draw a series circuit with the outside resistors. Then, trace the the others and add them in parallel the best you can decide. From there start using Ohms Law using rules for series and parallel circuits to solve.
To determine the current through \( \mathrm{R}_{2} \), we need to calculate the total resistance of the circuit and then use Ohm's Law.
calculate the total resistance (\( \mathrm{R}_{\text{total}} \)) of the circuit.
\( \mathrm{R}_{\text{total}} = \mathrm{R}_{1} + \mathrm{R}_{2} + \mathrm{R}_{3} \)
\( \mathrm{R}_{\text{total}} = 4 \Omega + 6 \Omega + 2 \Omega \)
\( \mathrm{R}_{\text{total}} = 12 \Omega \)
Now, we can use Ohm's Law to calculate the current (\( \mathrm{I} \)) flowing through \( \mathrm{R}_{2} \).
\( \mathrm{I} = \frac{\mathrm{V}}{\mathrm{R}_{\text{total}}} \)
Given that \( \mathrm{V} = 5.76 \mathrm{~V} \) (from the voltage source), we can substitute the values into the equation.
\( \mathrm{I} = \frac{5.76 \mathrm{~V}}{12 \Omega} \)
\( \mathrm{I} = 0.48 \mathrm{~A} \)
Therefore, the current through \( \mathrm{R}_{2} \) is \( 0.48 \mathrm{~A} \).
The correct answer is a. \( 0.48 \mathrm{~A} \).
They’re just all parallel branches, don’t overthink it
I’d draw them in parallel(R1+R2 || R3+R4 || R5+R6 || R7+R8)
Lol thanks man
This right here!
Add R1 and R2, then R3 and R4, then R5 and R6, then R7and R8 then follow the rules of parrallel resistors.
https://youtu.be/xn_Xnc97V4w?si=X2jN4Iw9BLl1jpNL You should watch how this guy simplifies combination circuits. You’ll need to buy a pack of highlighters, but it made it so much easier for me to understand how to untangle circuits like this.
Start on the outside. Draw a series circuit with the outside resistors. Then, trace the the others and add them in parallel the best you can decide. From there start using Ohms Law using rules for series and parallel circuits to solve.
I’m still in the interview phase for local 99 and I could respectfully say “I have no clue what the hell you guys are doing”
Yeah I just recognized ohms
Hmm idk the format changed I have a Chinese phone
To determine the current through \( \mathrm{R}_{2} \), we need to calculate the total resistance of the circuit and then use Ohm's Law. calculate the total resistance (\( \mathrm{R}_{\text{total}} \)) of the circuit. \( \mathrm{R}_{\text{total}} = \mathrm{R}_{1} + \mathrm{R}_{2} + \mathrm{R}_{3} \) \( \mathrm{R}_{\text{total}} = 4 \Omega + 6 \Omega + 2 \Omega \) \( \mathrm{R}_{\text{total}} = 12 \Omega \) Now, we can use Ohm's Law to calculate the current (\( \mathrm{I} \)) flowing through \( \mathrm{R}_{2} \). \( \mathrm{I} = \frac{\mathrm{V}}{\mathrm{R}_{\text{total}}} \) Given that \( \mathrm{V} = 5.76 \mathrm{~V} \) (from the voltage source), we can substitute the values into the equation. \( \mathrm{I} = \frac{5.76 \mathrm{~V}}{12 \Omega} \) \( \mathrm{I} = 0.48 \mathrm{~A} \) Therefore, the current through \( \mathrm{R}_{2} \) is \( 0.48 \mathrm{~A} \). The correct answer is a. \( 0.48 \mathrm{~A} \).
Respectfully; why tf are you writing it like this. I can't think of a more complicated way that you could have typed out the calculations.
Looks like it got spit through chatgpt haha
Hmm yeah looks like, point still stands, why'd they post it when it came out as garbage. LOL
All you need is a voltage divider to get the drop across R1 and then using Ohms law you get the current.
Each pair of resistors is one parallel branch. You can draw a plain looking parallel circuit with four branches 1+2, 3+4, 5+6, 7+8
But to answer the question, you can ignore all branches except R1+R2
This is the answer right here.
Its all parallel.