Potential Difference Across Resistors in Circuit

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SUMMARY

The discussion focuses on calculating the potential difference across a 3-ohm resistor in a DC circuit. The user initially calculated the current as 3 Amperes using the formula I = V/R. However, confusion arose when applying the second equation E = IR, leading to an incorrect voltage calculation. The correct approach involves determining the equivalent resistance (Req) as 2 ohms, which clarifies the voltage distribution across the resistors in series.

PREREQUISITES
  • Understanding of Ohm's Law (I = V/R)
  • Knowledge of series and parallel resistor configurations
  • Familiarity with equivalent resistance calculations
  • Basic principles of DC circuits
NEXT STEPS
  • Study the concept of equivalent resistance in series and parallel circuits
  • Learn how to apply Kirchhoff's Voltage Law in circuit analysis
  • Explore practical applications of Ohm's Law in real-world circuits
  • Investigate the effects of varying resistor values on circuit performance
USEFUL FOR

Electrical engineering students, hobbyists working with DC circuits, and anyone interested in understanding resistor behavior in electrical networks.

jensenns
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Almost forgot to mention:
This is a DC Circuit!

dumbsaus.jpg

In the diagram above, find the potential difference across the 3-ohm resistor.

a) 1.3 V
b) 3.0 V
c) 4.0 V
d) 5.0 V
e) 6.0 V

Relevant equations:
So far, I used: I = V/R to find current from the battery.
Then there's E = IR.
(Sorry, LaTeX doesn't like me.)

The attempt at a solution:
Using the first equation, I found that the current measures three Amperes.

I also tried using the second formula where I = 3 and R = 3, but that gave me 9 Volts, which isn't an answer.

Any help would be much appreciated!
 
Last edited:
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Welcome to PF.

Consider the equivalent resistance from the node above the 3Ω resistor to below.

If you determined the current was 3 a, then you must have determined that the Req was 2Ω.

In which case, your equivalent circuit looks like just two 2Ω resistors in series. That means then that the voltage at that node will be half of the battery supply.
 
Oh, duh! I totally forgot. Thank you!
 

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