Alternating Current Wheatstone-like Bridge

[ke7q]

Homework Statement

The goal of this problem is to find a frequency at which the voltage drop across several components of a circuit will be zero. The circuit is identical to a wheatstone bridge (where two pairs of two resistors in series are wired in parallel) except that opposite (diagonally) resistors are replaced with identical ideal inductors, and the other two resistors are replaced with identical ideal capacitors. Also, the voltage applied is not a DC voltage, and is instead given by

$$\epsilon = \epsilon_o Cos(\omega t)$$

Homework Equations

I know that the impedance of the capacitor and inductor are given by...

$$X_c = -\frac{i}{\omega C}$$

$$X_L = i \omega L$$

... and the reactance of these is just given by the magnitude of these in the imaginary plane.

The Attempt at a Solution

Even though the voltage drops across the inductors and capacitors are out of phase, my first thought was to use Kirchoff's rule and take a loop from one end of the current meter bridge to the other through one capacitor and one inductor. From this, I yield the equation that

$$Δ V_c = Δ V_L$$
$$-i\frac{I}{\omega C} = i \omega L I$$
$$-\frac{1}{\omega C} = \omega L$$

which yields that

$$\omega^{2} = -\frac{1}{LC}$$

But then I ran into a problem that I have a square of something that is supposed to be a real number being negative.

I considered just using the reactances from the beginning and just setting them equal, which yields the same as above, except for the minus sign, but I couldn't justify it.

Any ideas?

 

[ke7q]

And, of course, right after I post this I realize my problem.

My issue was that I was applying Kirchoff's loop rule incorrectly. Because I am going "against" the current in one case, I pick up a minus sign and didn't realize it. Therefore, the frequency I would want so that there is no voltage drop across the bridge would be

$$\omega = \frac{1}{\sqrt{LC}}$$

apologies for the spamish post on my very first post :/