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how to get to this KVL expression..

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nabliat6

New Member
in this photo there is the circuit and the KVL of the solution


but when i tried to do the KVL
i go from the arc(3 lines represents potential zero)
in counter clockwise direction
so i get[latex] V_L + V_C+V_R-V_s=0[/latex]
when i go against the direction of a current the voltage is on PLUS
when in the voltage source i go from + to - we have -V_s

so where is my mistake
i cant figure out how to get to the expression they got from my expression
?
 
Last edited:

nabliat6

New Member
how the solution could be current if the fracture represents current
on the resistor
(its a KVL)
does my logic in finding the KVL correct?
 
Last edited:

Electronworks

New Member
OK here goes...

Firstly, I would use the impedance of a capacitor as Z = 1/jxC and the impedance of an inductor as Z = jwL.

Using nodal analysis:

The current flowing in Vs, R and C is I1
The current flowing in L is I2

I1 = I2 + Is

therefore,

Vs + I1(R + 1/jwC) + I2(jwL) = 0
I1 - Is = I2

Therefore

Vs + I1(R + 1/jwC) + (I1-Is)(jwL) = 0

Is this what you were looking for...?
 
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