Find the Current Flowing through each resistor

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SUMMARY

This discussion focuses on calculating the current flowing through resistors in a circuit with batteries arranged in opposite polarity. Participants emphasize the importance of applying Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) to analyze the circuit effectively. The voltage drop across resistors is determined using the formula V=IR, where V represents voltage, I is current, and R is resistance. The conversation highlights that the assumed direction of current is crucial, as incorrect assumptions can lead to negative results in calculations.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Familiarity with Kirchhoff's Current Law (KCL)
  • Knowledge of Ohm's Law (V=IR)
  • Basic circuit analysis techniques
NEXT STEPS
  • Study advanced circuit analysis techniques using mesh and nodal analysis
  • Explore practical applications of Kirchhoff's Laws in real-world circuits
  • Learn about the impact of resistor configurations on current flow
  • Investigate the effects of battery polarity on circuit behavior
USEFUL FOR

Students studying electrical engineering, circuit designers, and anyone interested in mastering circuit analysis techniques.

Kelsi_Jade
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Homework Statement


upload_2017-3-27_18-32-38.png


Homework Equations

/ 3. The Attempt at a Solution
Looking to solve mistakes to study from homework. Here's what I had so far[/B]
upload_2017-3-27_18-32-1.png
 
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Good. Now: what is your question ?
 
The batteries are arranged in opposite polarity. Check your second equation.
 
I'm just wondering how to find the Voltage drop.
Would I need to find the voltage using V=IR for each of the currents that I found and then I could observe the drop from V1 to V2?
 
IS it clear what you mean with V1 and V2 ? What do you think of Magoo's remark ?
 
magoo said:
The batteries are arranged in opposite polarity. Check your second equation.
Aren't the currents going in the same direction, though? - They're both moving in a clock-wise direction?
 
Yes, so do the voltage drop over the 2 k##\Omega## and the 5V have he same sign or an opposite sign ?
 
Hmm...If you start with 10V and on the opposite side is 5V, is it bc of Kirchhoff's voltage law that the voltage drop would be negative over the 2kOhm to 5V battery?
Because your overall voltage of the loop should end up =0?
 
Kelsi_Jade said:
Because your overall voltage of the loop should end up =0?
Sum of potential changes (gains and drops) should be zero.

Kelsi_Jade said:
If you start with 10V and on the opposite side is 5V, is it bc of Kirchhoff's voltage law that the voltage drop would be negative over the 2kOhm to 5V battery
Usually, you can't predict that. You should assume the directions of currents first and solve the KVL and KCL equations. If the current direction assumed is incorrect, you'll get a minus sign in the answer.
 
  • #10
A few things:

Voltage polarity is due to potential difference.

Your equations don't seem to be loop (mesh) equations.

The direction to assign a current for analysis is arbitrary, as long as you don't swap it's direction mid analysis. If you guessed the correct direction, the result will be positive. Otherwise, the result will be negative. Compare your result, to the direction you assigned, and that will tell you which way current is flowing.
 

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