My physics professor used vector sum to find net voltage at any instant in the following $RL$ circuit and said that it is equal to vector sum of phasor vector of potential drop across Resistor and phasor vector of potential drop across the inductor, but I disagree because i performed the mathematical sum of the respective potential drops and saw that the approach used by professor is only true when $V$ is $V_0$ (i.e. maximum potential of source).

And Why there is a phase difference between phasor of potential drop across Resistor and $V$ net in the $RL$ circuit? (I have attached the notes of my professor.)

enter image description here


5 Answers 5


The quantities themselves are not vectors. But when we consider the general case, it can be proven and seen that their calculations are similar to the calculations to add vectors. That's why , it's convenient to establish short tricks ( here used to relate it to vectors) to avoid repeating a lengthy algorithm and to shorten the calculations.

Why there is a phase difference between phasor of potential drop across Resistor and $V$ net in the $RL$ circuit?

It's because the voltage across resistor is given by $RI$ where $I$ is instantaneous current. Since the current does have a phase difference with the voltage , so does the resistor.

As for why current has a phase difference with the voltage , you need to look to inductors. In fact , it's one of the definitions of inductance ($L$) that :

$$V = -L\frac{\mathrm dI}{\mathrm dt}$$

so that :

$$I(t)= \frac{-1}{L}\int_0^tV\mathrm dt$$

Since our $V$ (in our case where it's a trigonometric function like sine or cosine) , the current could be found by integration. And integration of simple sinusoidal or cosine functions interchanges them, so as to give a phase difference.

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    $\begingroup$ And why there is a phase difference between phasor of potential drop across Resistor and V net in the RL circuit ? $\endgroup$
    – user378646
    Commented Sep 24, 2023 at 15:19
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    $\begingroup$ If you ad sin functions with the same frequency but different phase , it is very convenient to use the definition of sin as projection of an arrow in a circle. Than its easier to add the arrows instead of the projection. Or you know complex numbers and the complex e. function as addition of trigonometric functions, than again the complex numbers are depicted as arrows. and is is convenient to add them up. otherwise it gets complicated to ad for example $sin(\omega t)+ sin(\omega t+\phi)$ but it is possible and gives the same results, just try it out. $\endgroup$
    – trula
    Commented Sep 24, 2023 at 15:20
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    $\begingroup$ @Phasor The derivations are easy and are usually covered in textbooks. I suggest you to read textbook(s) too along with lectures particularly for derivations. $\endgroup$ Commented Sep 24, 2023 at 15:28
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    $\begingroup$ @trula Sir I did that, but i am getting the result of vector addition and addition by mathematical tools same only when V is Vmax (see i have attached the calculation part) $\endgroup$
    – user378646
    Commented Sep 24, 2023 at 15:30
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    $\begingroup$ @An_Elephant can you recommend some books, i tried to verify with Resnick halliday but there the book is using spring mass system instead of phasor $\endgroup$
    – user378646
    Commented Sep 24, 2023 at 15:32

Let's work through the problem through 2 different but equivalent ways:

Trigonometric approach

In the first approach, we do the same calculation that you did. I'll use uppercase $V$ to denote the amplitude (maximum voltage) and lowercase $v(t)$ to denote the time-dependent voltage.

We start with the voltages of the resistor and the inductor:

$$ \begin{align*} v_{\text{R}}(t) &= V_{\text{R}} \sin(\omega t) \\ v_{\text{L}}(t) &= V_{\text{L}} \cos(\omega t) \end{align*} $$

and then just add them up:

$$ \begin{align*} v_{\text{net}}(t) &= v_{\text{R}}(t) + v_{\text{L}}(t) \\ &= V_{\text{R}} \sin(\omega t) + V_{\text{L}} \cos(\omega t) \end{align*}$$

So far so good: this is the same answer as your derivation.

If you look closer, you might notice the resemblance to a trigonometric identity:

$$a \cos x + b \sin x = c \cos(x + \varphi) \tag{I}$$

where $c = \operatorname{sgn}(a) \sqrt{a^2 + b^2}$ and $\varphi = \arctan(-b/a)$. Using that identity, we can rewrite the result:

$$ \begin{align*} v_{\text{net}}(t) &= V_{\text{R}} \sin(\omega t) + V_{\text{L}} \cos(\omega t) \\ &= \sqrt{V_{\text{R}}^2 + V_{\text{L}}^2} \cos\biggl(\omega t + \arctan\biggl(-\frac{V_{\text{R}}}{V_{\text{L}}}\biggr)\biggr) \end{align*} $$

In order to compare the phase of $v_{\text{net}}(t)$ to that of $v_{\text{R}}(t) = \sin(\omega t)$, we have to convert the cosine into a sine. Fortunately there are some more trigonometric identities that we can use:

$$ \begin{align*} \sin\biggl(\theta + \frac{\pi}{2}\biggr) &= \cos \theta \\ \tan\biggl(\theta + \frac{\pi}{2}\biggr) &= -\cot \theta \end{align*} \tag{II} $$

With that, we obtain the result:

$$ \begin{align*} v_{\text{net}}(t) &= \sqrt{V_{\text{R}}^2 + V_{\text{L}}^2} \cos\biggl(\omega t + \arctan\biggl(-\frac{V_{\text{R}}}{V_{\text{L}}}\biggr)\biggr) \\ &= \sqrt{V_{\text{R}}^2 + V_{\text{L}}^2} \sin\biggl(\omega t + \arctan\biggl(-\frac{V_{\text{R}}}{V_{\text{L}}}\biggr) + \frac{\pi}{2}\biggr) \\ &= \sqrt{V_{\text{R}}^2 + V_{\text{L}}^2} \sin\biggl(\omega t + \arctan \frac{V_{\text{L}}}{V_{\text{R}}}\biggr) \end{align*} $$

Now we can just read off the amplitude and phase of $v_{\text{net}}$:

$$ \begin{align*} \text{(amplitude of $v_{\text{net}}$)} &= \sqrt{V_{\text{R}}^2 + V_{\text{L}}^2} \\ \text{(phase of $v_{\text{net}}$ relative to $v_{\text{R}}$)} &= \arctan \frac{V_{\text{L}}}{V_{\text{R}}} \end{align*} $$

Phasor approach

The phasor approach comes from a careful observation of Euler's formula:

$$\mathrm{e}^{\mathrm{i} x} = \cos x + \mathrm{i} \sin x$$

If we project out the imaginary parts of both sides, we obtain:

$$\operatorname{Im} \bigl(\mathrm{e}^{\mathrm{i} x}\bigr) = \sin x \tag{IIIa}$$

Geometrically, $\mathrm{e}^{\mathrm{i} x}$ is a 2D unit vector that rotates counterclockwise as $x$ increases. When $x=0$, the unit vector points to +x and it completes a revolution every $2 \pi$ radians. The $\operatorname{Im}(\ldots)$ operation projects out the y-coordinate, which equals $\sin x$. (The geometric interpretation equivalent to the complex plane interpretation, where we associate the real axis with the x-axis and the imaginary axis with the y-axis.)

We can also get a similar relationship for cosine:

$$\operatorname{Im} \bigl(\mathrm{i} \mathrm{e}^{\mathrm{i} x}\bigr) = \cos x \tag{IIIb}$$

Now, we can transform the voltages using (IIIa) and (IIIb):

$$ \begin{align*} v_{\text{R}}(t) &= V_{\text{R}} \sin(\omega t) = V_{\text{R}} \operatorname{Im}\bigl(\mathrm{e}^{\mathrm{i} \omega t}\bigr) = \operatorname{Im}\bigl(V_{\text{R}} \mathrm{e}^{\mathrm{i} \omega t}\bigr) \\ v_{\text{L}}(t) &= V_{\text{L}} \cos(\omega t) = V_{\text{L}} \operatorname{Im}\bigl(\mathrm{i} \mathrm{e}^{\mathrm{i} \omega t}\bigr) = \operatorname{Im}\bigl(\mathrm{i} V_{\text{L}} \mathrm{e}^{\mathrm{i} \omega t}\bigr) \end{align*} $$

The expression inside $\operatorname{Im}\bigl(\cdots\bigr)$ is what is often called the "phasor". [Note: You can also define phasors using $\operatorname{Re}$ if your resistor had been $\propto \cos(\omega t)$.] We can give the phasors some names, for example:

$$ \begin{align*} v_{\text{R}}(t) &= \operatorname{Im}\bigl(u_{\text{R}}(t)\bigr) &\text{where} && u_{\text{R}}(t) &= V_{\text{R}} \mathrm{e}^{\mathrm{i} \omega t} \\ v_{\text{L}}(t) &= \operatorname{Im}\bigl(u_{\text{L}}(t)\bigr) &\text{where} && u_{\text{L}}(t) &= \mathrm{i} V_{\text{L}} \mathrm{e}^{\mathrm{i} \omega t} \end{align*} $$

Geometrically, $u_{\text{R}}(t)$ and $u_{\text{R}}(t)$ are both vectors rotating counterclockwise with angular frequency $\omega$. When $t = 0$, the $u_{\text{R}}(t)$ points to +x, whereas $u_{\text{L}}(t)$ points to +y. The length of $u_{\text{R}}(t)$ is $V_{\text{R}}$, whereas the length of $u_{\text{L}}(t)$ is $V_{\text{L}}$.

To find the net voltage, we need to add the two voltages together as before:

$$ \begin{align*} v_{\text{net}}(t) &= v_{\text{R}}(t) + v_{\text{L}}(t) \\ &= \operatorname{Im}\bigl(V_{\text{R}} \mathrm{e}^{\mathrm{i} \omega t}\bigr) + \operatorname{Im}\bigl(\mathrm{i} V_{\text{L}} \mathrm{e}^{\mathrm{i} \omega t}\bigr) \\ &= \operatorname{Im}\bigl((V_{\text{R}} + \mathrm{i} V_{\text{L}}) \mathrm{e}^{\mathrm{i} \omega t}\bigr) \end{align*} $$

As you can see, this is equivalent to just adding the phasors:

$$ \begin{align*} u_{\text{net}}(t) &= u_{\text{R}}(t) + u_{\text{L}}(t) \\ &= (V_{\text{R}} + \mathrm{i} V_{\text{L}}) \mathrm{e}^{\mathrm{i} \omega t} \end{align*} $$

where we define

$$v_{\text{net}}(t) = \operatorname{Im}\bigl(u_{\text{net}}(t)\bigr)$$

To find the amplitude and phase of a phasor, we have to convert it into polar form. There is a handy identity for this:

$$z = |z| \mathrm{e}^{\mathrm{i} \arg(z)}$$

With that, we can then further rewrite the phasor as:

$$ \begin{align*} u_{\text{net}}(t) &= (V_{\text{R}} + \mathrm{i} V_{\text{L}}) \mathrm{e}^{\mathrm{i} \omega t} \\ &= |V_{\text{R}} + \mathrm{i} V_{\text{L}}| \mathrm{e}^{\mathrm{i} \arg(V_{\text{R}} + \mathrm{i} V_{\text{L}})} \mathrm{e}^{\mathrm{i} \omega t} \\ &= \sqrt{V_{\text{R}}^2 + V_{\text{L}}^2} \mathrm{e}^{\mathrm{i} \arctan(V_{\text{L}}/V_{\text{R}})} \mathrm{e}^{\mathrm{i} \omega t} \\ &= \sqrt{V_{\text{R}}^2 + V_{\text{L}}^2} \mathrm{e}^{\mathrm{i} (\omega t + \arctan(V_{\text{L}}/V_{\text{R}}))} \end{align*} $$

Geometrically, this phasor $u_{\text{net}}(t)$ is a vector of length $\sqrt{V_{\text{R}}^2 + V_{\text{L}}^2}$ rotating with angular frequency $\omega$. At $t = 0$, the phasor starts at an angle of $\arctan (V_{\text{L}} / V_{\text{R}})$ counterclockwise from the +x-axis.

Now we can convert this back using (IIIa):

$$ \begin{align*} v_{\text{net}}(t) &= \operatorname{Im}\bigl(u_{\text{net}}(t)\bigr) \\ &= \operatorname{Im}\biggl(\sqrt{V_{\text{R}}^2 + V_{\text{L}}^2} \mathrm{e}^{\mathrm{i} (\omega t + \arctan(V_{\text{L}}/V_{\text{R}}))}\biggr) \\ &= \sqrt{V_{\text{R}}^2 + V_{\text{L}}^2} \operatorname{Im}\biggl(\mathrm{e}^{\mathrm{i} (\omega t + \arctan(V_{\text{L}}/V_{\text{R}}))}\biggr) \\ &= \sqrt{V_{\text{R}}^2 + V_{\text{L}}^2} \sin\biggl(\omega t + \arctan \frac{V_{\text{L}}}{V_{\text{R}}}\biggr) \end{align*} $$

which as you can see, yields at the same result as the trigonometric approach.

The main benefit of phasors / complex numbers is that the algebra is often much easier: rules are simpler and there are overall fewer rules to memorize. Contrast that with trigonometric functions, where you have be familiar with this enormous list of trigonometric identities!

  • $\begingroup$ Really nice. I didn't remember that the phasor was associated with the imaginary part, perhaps I just forgot, or perhaps my EE books used a different convention. But super work here!!! $\endgroup$ Commented Sep 27, 2023 at 14:48
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    $\begingroup$ @JosephDoggie "I didn't remember that the phasor was associated with the imaginary part," It can be either real or imaginary part depending on convention. I used imaginary part because it lines up with the OP's use of sin(ω t) for the resistor. $\endgroup$
    – Rufflewind
    Commented Sep 28, 2023 at 3:57

It's a concatenation of math tricks. We a complex analytic signal rather than a real-valued physical signal. This allows us to use the Fourier transform to reduce linear differential equations to complex algebraic equations: $\frac{d}{dt}e^{i\omega t}=i\omega e^{i\omega t}$ motivates the substitution $\frac{d}{dt}\rightarrow i\omega$. We use the vector representation of complex numbers in the complex plane to reason about the equations. Finally, we take the real part of the result to get a result representing the expected physics.

And once you're used to doing problems this way it's so easy and obvious that it's hard to explain.

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    $\begingroup$ So there exists a long derivation which proofs that vector sum can be employed to add the voltages ? $\endgroup$
    – user378646
    Commented Sep 24, 2023 at 15:58
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    $\begingroup$ @Phasor Well, I don't know how long it is. I sketched it in a paragraph. But derivations are for mathematicians: the way to gain confidence in this model is to make circuits and see what they do. $\endgroup$
    – John Doty
    Commented Sep 24, 2023 at 16:06

When we take any component in an AC circuit with Inductor, Capacitor, and Resistor, the mapping of the current/voltage into voltage/current would involve differential operators. For example, across an inductor we have the following mapping for voltage:

$$V_{\text{indtr}} : I_{\text{indtr}} \to - L \frac{dI_{\text{indtr}} }{dt}$$

By the nature of solutions/physical experiment to these simple AC circuits, we can find that the Inductor must necessarily have an expression of form in steady state:

$$I_{\text{indtr}} = Ae^{i \omega t}$$

This simplifies the expression of our mapping quite a bit:

$$V_{\text{indtr}} : I_{\text{indtr}} \to -i \omega L I_{\text{indtr}}$$

As a result, we find that the action involving differential operators on the current function can be seen as a complex multiplication.

Now that we have this, we apply the Kirchoff voltage law through the loop, eg, take the one below:

enter image description here

We would have:

$$V_{\text{btry} } = V_{\text{indtr}} + V_{\text{restr}}$$

Now, the teacher has used an additional fact to simplfy calculations, and that is, that the voltage for resistor and inductor should be perpendicular in the sense of the geometry present in al quantities belonging in $\mathbb{C}$. Hence, the vertical component of the battery voltage should be that of inductor, and of horizontal should be of resitor.

But, suppose we didn't have such a nice scenario, What if, for example, the Inductor's voltage was at an angle to the resistor?

Well, then we could do the law of cosine. For example, for the inductor, we have:

$$| V_{\text{indtr}}|^2 = |V_{\text{btry}}|^2 +| V_{\text{res}}|^2 - 2|V_{\text{btry}}|| V_{\text{res}}| \cos \theta$$

Where $\theta$ is the angle between Battery's vector and that of resistor.

Does this really work? Let's check if agrees with the case we had in the question. In question, we have $|V_{res}| = |V_{btry}| \cos \theta$, which leads us to have:

$$| V_{\text{indtr}}|^2 = |V_{\text{btry}}|^2 +|V_{\text{btry}}^2| \cos \theta^2 - 2|V_{\text{btry}}| |V_{\text{btry}}| \cos \theta \cos \theta= |V_{\text{btry}}|^2 - |V_{\text{btry}}|^2 \cos^2 \theta = |V_{\text{btry}}|^2 \left( 1- \cos^2 \theta\right) = |V_{\text{btry}}^2| \sin^2 \theta$$

Square rooting both sides, we see that, es stimmt alles.( everything works out)


If we allow magnitude and phase to vary, but keep frequency constant, we have two degrees of freedom, and the set of functions of the form $V(t) = A\sin(t-\phi)$ constitute a two-dimensional vector space. This space, the two-dimensional Cartesian plane, and the set of complex numbers, are isomorphic to each other with respect to addition (pointwise multiplication, the dot product, and complex multiplication, on the other hand, are all different operators). If we take $A\sin(t-\phi) \rightarrow [A\cos(\phi), A\sin(\phi)]$, then adding in the Cartesian plane corresponds to adding the functions.

That is, since $A\sin(t-\phi) = A(\sin t \cos \phi -\sin \phi \cos t)$, the transformation $A\sin(t-\phi) \rightarrow [A\cos(\phi), A\sin(\phi)]$ corresponds to taking $x$ to be coefficient of $\sin t$ and $y$ to be the negative of the coefficient of $\cos t$. If we have $V_1(t) = A_1\sin(t-\phi_1)$ and $V_2(t)=A_2\sin(t-\phi_2)$, taking them to the Cartesian plane and then adding them together results in $[A_1\cos(\phi_1)+A_2\cos(\phi_2), A_1\sin(\phi_1)+A_2\sin(\phi_2)]$. If you were to instead add the functions directly, you would get $V_1(t)+V_2(t)=A_1(\sin t\cos\phi_1-\sin\phi_1\cos t)+A_2(\sin t\cos\phi_2-\sin\phi_2\cos t)=(A_1\cos\phi_1+A_2\cos\phi_2)\sin t+(A_1\sin\phi_1+A_2\sin\phi_2)\cos t$. So $A_1\cos\phi_1+A_2\cos\phi_2$ is indeed the coefficient of $\sin t$ and $A_1\sin\phi_1+A_2\sin\phi_2$ the coefficient of $\cos t$.

Also, while $V_0$ is pronounced the same as "vee not", it should be written as "V_0$ or $V naught". "Naught" is another word for "zero".


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