 #1
Jaccobtw
 162
 31
 Homework Statement:
 In an RLC series circuit that includes a source of alternating current operating at a fixed frequency and voltage, the resistance of 10Ω is equal to the inductive reactance. If the capacitor plate separation is reduced to one half its original value, the magnitude of the current doubles in the circuit. Find the initial capacitive reactance in Ω. Note that although we adjust the separation, you can still treat the capacitor as an ideal parallel plate capacitor.
 Relevant Equations:

$$Capacitive reactance = X_C = \frac{1}{\omega C} = \frac{V_0}{I_0}$$
$$Inductive Reactance = X_L = \omega L = \frac{V_0}{I_0}$$
I'm just confused about how to go about this problem. I don't have all variables. What does the current doubling with half plate separation have to do with the problem? Your help is much appreciated. Thank you.