What does it actually mean to add voltages vectorially not algebraically in RL or RC circuits (AC source)? I know that the voltage drop across the inductor is maximum when the voltage drop through the resistor is zero, then VR continues to increase and VL decreases, so why can't we simply add them when we calculate the total voltage drop across the two? what is the secret behind using vectors here?

  • $\begingroup$ To be clear, are you asking about adding time domain voltages (point wise) or adding the phasor domain voltages? Typically, AC circuits are analyzed in the phasor domain where the (phasor) voltages are complex constants. But, in your question, you're talking about time varying voltages so you must be thinking time domain? But in the time domain, voltages are simply added together. Please clarify what precisely you're asking. $\endgroup$ Mar 9 '19 at 20:40

so why can't we simply add them when we calculate the total voltage drop across the two?

You do simply add the time domain voltages, at each moment in time, together. For example, stipulate that the resistor and inductor are series connected, and that the series current through, $i(t)$, is of the form

$$i(t) = I_0\sin(\omega t)$$

The voltage across the series combination is simply the sum of the voltages across the individual circuit elements

$$v(t) = RI_0\sin(\omega t) + \omega LI_0\cos(\omega t)$$

Now, you may know that you can write this as a single sinusoidal function:

$$v(t) = \sqrt{(RI_0)^2 + (\omega L I_0)^2}\,\sin(\omega t + \phi),\quad \tan\phi = \frac{\omega L}{R}$$

Notice that the amplitude of this sinusoid is found by adding the amplitudes of the individual sine and cosine in quadrature (like finding the length of vector).

What is the secret behind using vectors here?

The result above is a hint. In the simple sum expression for $v(t)$, one term is a sine while the other is a cosine. These functions are orthogonal and so we can think of the sum as being the sum of two orthogonal 'vectors'. The result is a 'vector' where each term is a separate 'component'.

Recall that a plane vector has a Cartesian form

$$\vec v = (x, y)$$

as well as a polar form

$$\vec v = (r, \phi)$$


$$r = \sqrt{x^2 + y^2},\quad \tan\phi = \frac{y}{x}$$

So the second expression for $v(t)$ can be thought of as a kind of 'polar form' of $v(t)$.

This type of thinking comes into clear focus when we switch from the time domain to the phasor domain where each voltage and current is a complex number that expresses the amplitude and phase only of the sinusoidal function of time. Since another answer addresses your question in the context of AC phasor analysis, I'll not go further into that here.


Hyportnex has given you an explanation in the time domain. The following involves phasor notation, which is in the frequency domain.

You need to add ac voltages vectorially for RC and LC circuits because the voltage and currents in inductors and capacitors are 90 degrees out of phase with each other, whereas the voltage and current are in phase for a resistor. This is a consequence of the following voltage current relationships for capacitors and inductors. $$i_{C}(t)=C\frac {dV_{C}(t)}{dt}$$ $$v_{L}(t)=L\frac {di_{L}(t)}{dt}$$ Note that if the voltage across the capacitor is a sine wave, the current will be the derivative of the sine wave, or cosine, which is $90^0$ out of phase with the sine. Similarly, if the current in an inductor is a sine wave, the voltage across the inductor will be a cosine. For the capacitor, the current leads the voltage by $90^0$. For an inductor the current lags the voltage by $90^0$. For a resistor the voltage and current are in phase (zero phase angle)

The impedances for capacitors, inductors, and resistors can be written using phasor notations. $$Z_{C}=X_{C}\angle -90^0$$ $$Z_{L}=X_{L}\angle +90^0$$ $$Z_{R}=R\angle 0^0$$ where $X_C$ and $X_L$ are the capacitive and inductive reactances given by $$X_{C}=\frac {1}{2πfC}$$ $$X_{L}=2πfL$$

As an example, if we have a series RC circuit with a phasor current of $I\angle 0^0$, the voltages across the capacitor and resistor will be, in phasor, notation

$$V_{C}=IZ_{C}=(I\angle 0^0)(X_{C}\angle -90^0) = IX_{C}\angle-90^0$$ $$V_{R}=IZ_{R}=(I\angle 0^0)(R\angle 0^0) = IR\angle 0^0$$

Since $V_R$ and $V_C$ are $90^0$ out of phase with each other, if you want to add them you will have to be added vectorially. You can get the magnitude of the complex voltage by taking the square root of the sum of the squares of the voltage magnitudes. You get the resulting phase angle between the real and imaginary components by taking the inverse tangent of $\frac {-X_{C}}{R}$.

Hope this helps.

  • $\begingroup$ Just a pedantic note FWIW: many (myself included) write $Z = R + jX$, i.e., that reactance is the imaginary part of impedance, and it then follows that $X_C = -\frac{1}{2\pi f C}$ so that $Z_C = jX_C$ just as $Z_L = jX_L$. Others will write, as you have, that $X_C = \frac{1}{2\pi fC}$ but it then must be that $Z_C = -jX_C$ which conflicts with $Z = R + jX$. These 'dueling' conventions are discussed at the Wikipedia article here. $\endgroup$ Mar 10 '19 at 1:55
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    $\begingroup$ @AlfredCentauri Yes, I'm aware of the fact that sometimes the sign (+/-) is assigned to the reactance and sometimes it is assigned to the imaginary number j. I was taught the latter (> 50 years ago) from the standpoint that reactance is the magnitude but no the sense. But it all comes out the same in the wash. Thanks for pointing it out. $\endgroup$
    – Bob D
    Mar 10 '19 at 2:03

You are adding two sinusoidal oscillations of the same frequency but differing in phase and amplitudes, say $V_1 (t)= A_1 \cos(\omega t +\phi_1)$ and $V_2(t) = A_2 \cos(\omega t +\phi_2)$. Now write this as $$V_1 +V_2 = A_1 \cos(\omega t +\phi_1) + A_2 \cos(\omega t +\phi_2)\\ =(A_1 \cos(\phi_1) + A_2 \cos(\phi_2)) \cos(\omega t) - (A_1 \sin(\phi_1) + A_2 \sin(\phi_2)) \sin(\omega t).$$ Set $B = A_1 \cos(\phi_1) + A_2 \cos(\phi_2)$ and $C= -A_1 \sin(\phi_1) - A_2 \sin(\phi_2)$ and notice that when you add the cartesian vectors $\textbf{v}_1 = [A_1 \cos(\phi_1), -A_1 \sin(\phi_1)]$ and $\textbf{v}_2 = [A_2 \cos(\phi_2), -A_2 \sin(\phi_2)]$ their sum has the same coordinates as $\textbf{v}_1+\textbf{v}_2=[B,C]$


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