# what is the magnitude of total voltage in the below circuit?

I was working on some RLC circuits and I found a question something like this.

In series RL circuit, $V_R$ = 4V and $V_L$ = 3V. what is the magnitude of total voltage ?

I just thought $V_T$ = 4V + 3V = 7V but the book says its 5V. Could anyone explain me am I right or wrong in thinking so. Is it 5V or 7V ? Please help me with these.

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It will only be 7V in first grade. In any later year you should know there will be something strange happening. –  Federico Russo May 21 '12 at 9:35

Whenever you have a resistor in series with an (ideal) inductor, if the current is sinusoidal, their voltages will be 90º apart. The total voltage $V_T$ is the vector sum of $V_R$ and $V_L$. Since $V_R$ and $V_L$ are orthogonal (due to the 90º phase difference), the module of $V_T$ can be easily computed as $|V_T|=\sqrt{|V_R|^2+|V_L|^2}=\sqrt{4^2+3^2}=\sqrt{25}=5$.

The reason for those 90º is: if $I=\sin(wt)$, then it will be $V_R=R\sin(wt)$ and $V_L=L\dfrac{dI}{dt}=L·w\cos(wt)$ .

Update: stevenh is right, and my explanation could be confusing. I was indeed assuming, from the OP, some two-dimensional interpretation of sinusoidal voltages and currents (where A·sin(wt+phi) is just a rotating vector with radius A and phase wt+phi), although complex numbers are not really required.

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@Telaclavo: "although complex numbers are not really required". The rotating vector will be rotating in the complex Argand plane, with a real axis and an imaginary axis. It's the basic representation of complex numbers. And if the inductor wouldn't have a complex impedance you would never have the 90 degree phase shift. I can't see how you can do this without complex numbers. –  Federico Russo May 21 '12 at 13:44
@FedericoRusso If A·sin(wt+phi) is a rotating vector with radius A and phase wt+phi, B·cos(wt+phi), being equal to B·sin(wt+phi+pi/2), is a rotating vector with radius B and phase wt+phi+pi/2. Since the phase difference is pi/2, those two vectors are orthogonal, and I can use the formula I used. No need to use complex numbers, or mention "j". // Vectors don't involve complex numbers. // And $V_L=L\dfrac{dI}{dt}$ doesn't involve complex numbers. –  Telaclavo May 21 '12 at 15:32
@Telaclavo: I think I understand what you're saying, but the sine can only rotate due to $sin(x)= \dfrac{e^{i x}-e^{-i x}}{2i}$, right? –  Federico Russo May 21 '12 at 15:44
@federico - No. $e^{j\omega t}$ and $e^{-j\omega t}$ are phasors which rotate in opposite directions. Their difference is a vector which grows and shrinks, but is always on the imaginary axis. Dividing by $j$ rotates this 90° clockwise, so it moves to the real axis. Sin(x) is a scalar, not a phasor. Like so many I learned this by heart at school, but you don't have to if you understand what the equation represents; you can always reconstruct it. –  stevenvh May 22 '12 at 3:49

The right answer is 5V as the others already explained. I'll assume you have a sinusoidal signal applied (otherwise you would have got a different result).

Impedance of the resistor is $R$, which is a real value.

Impedance of the inductor is $j\omega L$, which is complex. While multiplying by a real value scales a vector, multiplying by (a power of) $j$ rotates it in the complex plane. Multiplying by $j$ gives a 90° rotation, $\times j^3$ is 3 $\times$ 90°, and $\times \sqrt{j}$ will give a 45° rotation, for instance.

(In the image $i$ is used instead of $j$. That's what mathematicians use. In electronics $j$ was chosen because $i$ was already used to indicate current.)

$V_R = I \times R$

Voltage and current have the same phase; their vectors point in the same direction.

$V_L = I \times j\omega L$

The factor $j$ causes a 90° rotation of the $I$ vector, so the voltage is at a right angle.
Now $I$ is the same for resistor and inductor since they're in series. $V_R$ is in phase with $I$, and $V_L$ is at 90° with that same $I$, therefore $V_L$ and $V_R$ are at a right angle. Adding them gives you a right-angle triangle, and you can apply Pythagoras to find the magnitude of the sum:

$|V| = \sqrt{|V_L|^2 +|V_R|^2} = \sqrt{(3V)^2 +(4V)^2} = 5V$

The phase difference between current and voltage is

$\phi = arctan\left(\dfrac{V_L}{V_R}\right) = arctan\left(\dfrac{3V}{4V}\right) = 37°$

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$$V_{combined} = \sqrt{ Vr^2 + Vl^2} = \sqrt{3^2 + 4^2} = \sqrt {9 + 16} = \sqrt{25} = 5 \mathrm{ \, V}$$