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Homework Statement
Consider the following circuit where i(t) is sinusoidal and exists across both components. (1) is an inductor and (2) is a capacitor. The ideal voltmeters measure effective value. What is the value measured by V2:
[![enter image description here][1]][1] [1]: https://i.stack.imgur.com/d7OfJ.png
The correct answer should be 2V
Homework Equations
3. The Attempt at a Solution [/B]
My attempt.
I know we can only apply KVL using phasors. The current phasor should be $$Ie^{j\theta}$$
The voltage across the inductor is $$\omega LIe^{j(\theta+\pi/2)}=V_1e^{j(\theta+\pi/2)}$$
The voltage across the capacitor is $$\frac{I}{\omega C}e^{j(\theta+-\pi/2)}=V_2e^{j(\theta-\pi/2)}$$
And the voltage across both components $$(\omega L-\frac{1}{\omega C})Ie^{j(\theta+-\pi/2)}=Ve^{j(\theta+\pi/2)}$$
And so applying KVL we get to
$$Ve^{j(\theta+\pi/2)}=V_1e^{j(\theta+\pi/2)}+V_2e^{j(\theta-\pi/2)}$$
$$Ve^{j\pi/2)}=V_1e^{j(\pi/2)}+V_2e^{j(-\pi/2)}$$
And dividing everything by $$\sqrt{2}$$
$$V_{eff}e^{j\pi/2)}=V_{1eff}e^{j(\pi/2)}+V_{2eff}e^{j(-\pi/2)}$$
Substituting by the known values
$$1e^{j\pi/2)}=1e^{j(\pi/2)}+V_{2eff}e^{j(-\pi/2)}$$
$$0=V_{2eff}e^{j(-\pi/2)}$$
And so $$V_{2eff}=0$$.
However the correct answer should be +2V.
Can someone help me to figure out my mistake? It's probably a conceptual mistake. Thanks!
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