Circuit Problems : Finding Current and Voltage

AI Thread Summary
The discussion revolves around solving a circuit problem using Kirchhoff's Voltage Law (KVL) and Ohm's Law to find voltage (V) and current (i). The initial equations set up by the user included an incorrect term, "-V," which caused confusion in the calculations. After reevaluating the KVL path, it was clarified that the term was unnecessary, leading to a more accurate approach. The user initially calculated a current of 0.007A and a voltage of 42V, but this did not align with the expected range of 78V to 90V provided by the professor. Ultimately, correcting the KVL path helped resolve the issue and clarify the calculations.
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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