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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

3 Answers

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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
up vote 1 down vote

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.

enter image description here

(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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up vote 0 down vote

You need to provide far more information and a circuit or diagram.

BUT you are probably dealing with a circuit with resistive and reactive components at right angles. Vr is probably the resistive component and Vl = inductive component at 90 degrees. The resultant is the vector combination of the two = the hypotenuse of the triangle that Vr and Vl form the sides of.

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

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