Find the voltage at point X on the circuit

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SUMMARY

The voltage at point X in the circuit can be determined using Kirchhoff's law and the known values of the resistors and the battery. The calculated equivalent resistance (Req) of the circuit is 2.039 ohms, and the current values are I1 = -39.23A and I2 = -25.38A. The voltage at point X is found to be 78.46V or 1.54V, depending on the reference point used. The voltage difference is derived from the electromotive force (emf) of the 80V battery and the potential difference across the 2 ohm resistor.

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  • Basic concepts of voltage and current in electrical circuits
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mathman44
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Homework Statement



Here is the circuit:
Loop.jpg


I have found I1 to be -39.23A, I2 to be -25.38A, but I'm having difficulty reasoning how to find the voltage of point X.

The Attempt at a Solution



Here is what I tried. I calculated Req for the entire system by V/I = 2.039. Since there is already a 2 ohm resistor in the first loop, the system can be reduced to one loop with the 2 ohm resistor and a 0.039 resistor in place on the 3 ohm. Now "X" is between this 0.039 resistor and the 2 resistor, so using Kirchoff's law I found that X has a voltage of 78.46 or 1.54. Does this seem logical? I have to admit I'm grasping for straws here.
 
Last edited:
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The "earth" symbol at bottom left means that point in the circuit is taken as zero volts for reference purposes.
The voltage at X is simply, then, the pd between X and the Earth point.
One way you can find it is from the emf of the 80V battery and the pd across the 2 ohm resistor.
 
Last edited:
Thanks, that sounds like a much simpler way of doing what I did!
 

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