How can I find Voltage V in an AC circuit using loop rules?

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MechEngrStdnt
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AC circuit schematic with two loop currents i1 and i2 and node voltage V: a 20∠90° source with a 10 Ω resistor, a 10J Ω inductor down from node V, a -5J Ω capacitor, and a 10∠0° source on the right.

Homework Statement



Need to find Voltage V. I know that I need to first solve for the currents, have tried using loop rules, not having much luck.

Homework Equations



Just would like to know how to approach, and solve, this problem. ME and circuits has never been my friend, thanks!

The Attempt at a Solution


Handwritten nodal analysis working on lined paper solving for V1 with complex admittances, ending in V1 = 44.72 at 63.4 degrees in polar form.
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  • Handwritten derivation, rotated sideways, solving for node voltage V1 in an AC circuit: (V1 - 20J)/10 + V1/(10J) + (V1 - 10)/(-5J) = 0, simplified step by step to V1 = 40J - 20, then |V1| = 44.72 and angle 63.4 degrees.
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You chose an excellent strategy using nodal analysis as it yields an equation in one variable that happens to be the one you want to find.

Something went wrong between your third and fourth step though, when you tried to invert the coefficient of V1 in moving it to the RHS of the equation.
 
Handwritten whiteboard working of nodal analysis with complex admittances, using J for the imaginary unit, solving for V1 through successive algebraic steps ending in V1 = 4J(10 - 10J) = 40 + 40J.

Dealing with i has been messing me up, don't use it much in Newtonian mechanics haha. I think I'm getting further away... according to chegg the solution is 28.28<45deg. ill attach my latest attempt. Thanks for the help!
 
Yeah, you've done the same thing again between the final line on the left and the new line at the top right. Note:

[tex]\left( \frac{1}{10} + \frac{j}{10} \right)^{-1} \neq \left( 10 - 10j \right)[/tex]
 
Wow, embarrassing...

You sir are a scholar, and gentleman. Thanks!