# What does vector addition of voltages mean?

So I was studying LCR Circuits in Alternating Current and I found something pretty weird. We are treating the voltage as a vector (phasor) and then vectorially adding them to get the net voltage.

But this thing doesn't make sense to me: What does it mean to add voltages which differ in phase?

• Calculate $v_3(t) = V_1\sin(\omega t) + V_2\sin(\omega t + \phi)$. Check if it's the same thing you'd get by adding phasors. Jul 6 '18 at 19:50
• @ThePhoton Mathematically it is the same. I want to know what does this intuitively mean. Jul 6 '18 at 19:55
• It means adding the two v(t) functions instantaneously, and expressing the result as a phasor. Jul 6 '18 at 19:59
• Unlike DC signals, AC signals change with time and since two signals may not be the same frequency and not necessarily synchronized with one another, you have to take phase into account as well as magnitude. Two signals, same frequency, same amplitude, at one extreme (0 phase difference) you double the size. At another extreme (180 degree phase difference) they cancel one another. Jul 6 '18 at 20:12
• A phasor is an abstractions rather than a concrete thing. That is a voltage phasor is not a voltage. It's a calculating tool. A way to avoid doing lots of grotty trig identities (and as such it's a calculating tool that I very much appreciate). Jul 7 '18 at 3:17

If you study the graphs above you will see that the voltage across the capacitor lags behind the voltage across the resistor by a quarter of a period which is equivalent to a phase angle of $-90^\circ$ and the voltage across the inductor leads the voltage across the resistor by a quarter of a period which is equivalent to a phase angle of $+90^\circ$.
So to find the voltage across all three of the components $v_{\rm series}$ at any instant of time one has to evaluate $v_{\rm R} \sin (\omega t) + v_{\rm C} \sin (\omega t - 90^\circ) + v_{\rm L} \sin (\omega t + 90^\circ)$.
A convenient way to do this addition is to use a phasor diagram as shown on the right and you will see in this case $v_{\rm supply}$ leads the voltage across the resistor (and hence the current which is always in phase with the voltage across the resistor by a value which is between $0^\circ$ and $+90^\circ$.