How Do You Calculate the Voltage Across v1 Using KCL and KVL?

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SUMMARY

The discussion focuses on calculating the voltage across v1 in a circuit using Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL). The user initially calculated v1 as 3V, which was incorrect based on the provided options: 9V, -9V, -5V, and 4V. The correct approach involves recognizing the voltage across the current source and the voltage drop across the resistors, particularly the 1Ω resistor with a 3A current. The voltage across v1 can be determined by subtracting the voltage drop across the 1Ω resistor from the known voltage of 4V.

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  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Familiarity with Kirchhoff's Current Law (KCL)
  • Basic knowledge of Ohm's Law and resistor calculations
  • Ability to analyze simple electrical circuits
NEXT STEPS
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  • Learn about Thevenin's and Norton's Theorems for circuit simplification
  • Explore the impact of different resistor configurations on voltage and current
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Electrical engineering students, circuit designers, and anyone involved in analyzing electrical circuits using KVL and KCL.

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Homework Statement


Find the voltage across v1 in the circuit below:

http://puu.sh/1eD2R

Homework Equations


The Attempt at a Solution



I tried using KVL across loop 1 (from left) and then:
LOOP2: v1, 2ohms,1ohm,4V
LOOP3: also: 5V, 2ohms, 1ohm, 4V

I seem to be getting 3V as the voltage across v1. I don't think I am right because it's not one of the options. The options are:

a) 9 V
b) -9 V
c) -5 V
d) 4 V

My rough work, (if you can decipher my penmanship - near the end, I skipped a steps and just did it in my head/calculator's memory):
http://puu.sh/1eD7L

Also, is it possible to use the loop with the 3A current source to do KVL? I thought it wasn't allowed because I don't know the voltage drop across the current source.

Thanks!
 
Last edited by a moderator:
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You do know the voltage across the current source. It's 4V minus the drop across that right-most 2Ω resistor (that has 3A through it).

But a simpler way of looking at this is to say the voltage between the top and bottom horiz wires (on your left) is 4V minus the drop across the 1Ω (with 3A in it).
 

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