Potential Difference between two points of a circuit

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SUMMARY

The discussion focuses on calculating the potential difference (Vab) between two points in a circuit using Kirchhoff's Loop Law and Ohm's Law. The solution involves determining the current through the circuit, which is calculated as I = 45 V / 155 Ω, resulting in approximately 0.29 Amps. The potential difference is derived step-by-step, accounting for voltage drops across resistors and contributions from power supplies, ultimately concluding that Vab equals 11.95 V.

PREREQUISITES
  • Understanding of Kirchhoff's Loop Law
  • Proficiency in Ohm's Law
  • Familiarity with circuit analysis techniques
  • Basic knowledge of voltage, current, and resistance
NEXT STEPS
  • Study advanced applications of Kirchhoff's Laws in complex circuits
  • Learn about Thevenin's and Norton's Theorems for circuit simplification
  • Explore AC circuit analysis techniques
  • Investigate the impact of non-ideal components on circuit behavior
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing or designing electrical circuits will benefit from this discussion.

lylos
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Homework Statement


Given the following circuit, what is Vab? Explain...

circuit.jpg



Homework Equations


Kirchoff's Law and Ohm's Law


The Attempt at a Solution



Here is my solution... Is this describing it correctly?

Kirchoff’s Loop Law to find current:
-10 Ω (I) + 30 Ω V – 100 Ω (I) – 5 Ω (I) + 15 V – 40 Ω (I) = 0
45 V = 155 Ω (I)
I = 45 V / 155 Ω or .29 Amps

Potential Difference between points a and b:
1. At point a, we will assume there to be zero potential.
2. By going over the 5 Ω resistor, we lose 1.45 V (5 Ω * .29 A), thus leaving us at - 1.45 V.
3. The 15 V power supply adds 15 V, leaving us with 13.55 V.
4. The 40 Ω resistor takes away another 11.6 V (40 Ω * .29 A), leaving us with 1.95 V.
5. The 5 Ω resistor that we come in contact with next doesn’t take anything away as it doesn’t have current running through it, so we’re left with 1.95 V.
6. Now, we add another 10 V to the potential difference between the point a and point b due to the power supply.
7. We now have a combined Potential Difference between point a and point b to be 11.95 V.
 
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I didnt check your math but the rest seems good to me.
 

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