# How to calculate the phase shift between the voltage and current in a RC circuit

The question asks: A voltage of the form $V=V_0 sin (ωt)$ is applied across the network. Evaluate the phase shift between the voltage and the current flowing through the network. State whether the current leads or lags. I've found the complex impedance - it is $Z=R + 1/jwC$ where j is the imaginary unit, w is the angular frequency and C is the capacitance. I then found the modulus and angle of this impedance. However, I don't understand how I can relate this to current and find the phase difference.

Phasors confuse me as I'm not sure whether I should be taking the length of the voltage vector into account when calculating the angles. The complex exponential method, representing V as $V_0e^{j(wt)} = Z *I_0e^{j(wt+\phi)}$ confuses me as I don't know how to find Phi, with two unknowns $I_0$ and $V_0$ there.

My ultimate issue is that I'm receiving differing information from different sources. Some youtube guides are suggesting things different to my textbook. Guidance would be appreciated.

$V_0e^{j(wt)} = Z *I_0e^{j(wt+\phi)} = Z\times I_0 \times e^{j\omega t}\times e^{j\phi}$
$Z = \dfrac{V_0}{I_0}\times e^{-j\phi}=\dfrac{V_0}{I_0}\:\: \large \angle \small (-\phi)$
Now express your original expression for $Z$ in magnitude and angle form:
$Z=\small \sqrt {R^2+(1/\omega C)^2} \:\:\large\angle \small arctan(\frac{-1}{\omega RC})$
• $I_0=\frac{V_0}{\sqrt{R^2+(1/\omega C)^2}}$, at a phase angle = $arctan\frac{1}{\omega RC}$ relative to $V_0$ – Chu Jan 5 '16 at 18:10