Circuit Problems : Finding Current and Voltage

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SUMMARY

The discussion focuses on solving a circuit problem using Kirchhoff's Voltage Law (KVL) and Ohm's Law. The user initially derived equations based on the circuit's components, including a 72V source and resistors of 2000Ω and 3000Ω. The user calculated the current as 0.007A but encountered discrepancies in the voltage calculations, specifically for V, which should range from 78V to 90V. Clarification was provided regarding the inclusion of an extraneous term "-V," in the KVL equation, which was determined to be unnecessary.

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  • Understanding of Kirchhoff's Voltage Law (KVL)
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gatsbycollege
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Homework Statement


The Circuit is given as shown in the figure
http://img215.imageshack.us/img215/8361/circuitsl.jpg



find V,
V, = ranging from 78v - 90v

Homework Equations


KVL
E=E1+E2+E3...+En
KCL
I=I1=I2
Ohms Law
V=IR

The Attempt at a Solution


i used KVL
72v-2000ohms(i)-0.5V,-V,=0
then i have this equation
72v=1.5V, +2000i (1)

in the 2nd resistor with 3000ohms
i derived the equation
V,=(2I)(3000ohms)
so V,=6000ohmsi (2)

substitution in eq (2) and (1)

i=0.007A

then going back to eq (2)
V,=42v <---- but this is not correct by looking at the range my prof gave me..
so i think i missed something here or my whole idea is not correct..
btw i don't really know if that 2i is in series with the resistor 3000ohms
 
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gatsbycollege said:

The Attempt at a Solution


i used KVL
72v-2000ohms(i)-0.5V,-V,=0

Where does the extra "-V," come from at the end? When I follow a KVL path around the first loop I go through the 72V source, the 2000Ω resistor, then the "0.5V," source and then back to the beginning via path AB.
 
gneill said:
Where does the extra "-V," come from at the end? When I follow a KVL path around the first loop I go through the 72V source, the 2000Ω resistor, then the "0.5V," source and then back to the beginning via path AB.

whew that solve my problem.. that "-V," is not included XD
 

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