RL circuit

Because you do not add the figures up arithmetically. You must use vector addition. IF VR = VD across R and VL = VD across the reactor, then the voltage drop across the combination of R and XL = sqrt(IR² + IXL²). If I = 10 amps and R = 8 and XL = 6 ohms the the VD = sqrt[(10 x 8)² + (10 x 6)²] = 100 volts. You can check this as follows : Z = sqrt(R² + XL²) = sqrt(8² + 6²) = 10 ohms.

I = voltage/Z = 100 volts/10 ohms = 10 amps.

This "extra" voltage comes from the rate-of-change in the current through L, the inductor.

In a series RL circuit the same current flows through both components. The voltage across the resistor is proportional to the current through it. As you know, this ratio is expressed as Ω = E/I. Good 'ol Ohm's Law.

But the voltage across an inductor is not proportional to the current through it. The voltage is proportional to the rate of change of current through the inductor. In other words E_L = L di/dt, where:

E_L = The instantaneous voltage across the inductor

L = the inductance in Henrys

di/dt = the instantaneous rate of change in the current through the inductor.

For example, let the current in a series RL circuit increase linearly over time. A "current ramp" if you will. In other words, I = Kt, where K represents some current, and t is time in seconds. Together Kt represents a rate of change in current, because the current is changing over time.

Give K a value of, say, 2 amps. So, after one second the current is two amps. After two seconds it is four amps and so forth.

K = 2,

Now, if i = Kt, then di/dt = K (a little calculus here). In other words, di/dt = 2

Let's also give L a value of, say, 10 Henrys

The voltage across the inductor is then:

V = 2 A/s * 10 H = 20 volts. A constant 20 volts.

If we increase K to, say, 20 amps, the voltage across the inductor will increase as well, to:

V = 20 A/s * 10 H = 200 volts across the inductor. A constant 200 volts.

Notice I didn't even bother mentioning the resistor, and the voltage in this example has nothing to do with the resistance value at all, but only with the rate of change in current!

--

Now, instead of using a constantly changing current through the inductor, let's instead fix the current to some constant value: i = M amps. Not i = mt, but just i = M! Let's give M a value of 5 amps. What then is the value of di/dt?

di/dt is zero!

Of course it's zero! The current is not changing with time. It is, simply, 5 amps. Since the inductor only responds to a change in current, what then is the voltage across the inductor?

Zero!

Plain 'ol zero.

But there may be, say, 10 volts still across the resistor and, in this case, the applied voltage must be 10 volts as well since the inductor is contributing nothing. (The resistor in this case would have a value of 2 ohms, since R = E/I = 10/5 = 2 ohms).

Do you see how the voltage across the inductance can have a value irrespective of the circuit's applied voltage? The inductor's voltage is non-zero only when the current is changing. If it is changing rapidly, there will be a large voltage across the inductor. If a large current is flowing through the inductor and you suddenly open the switch, the current is immediately interrupted - an an extremely rapid change in current - so the inductor's voltage will try go through the roof (even though the applied voltage across the RL circuit as a whole may actually be quite low). That's why opened switch contacts arc when the load is highly inductive. There may be thousands of volts across the switch even though the circuit's applied voltage is not even close to that.

I hope this reply makes some sense.

Kind regards,

TV

In addition to the reason given by wareagle, you need to consider that every inductor has some stray capacitance. At some frequencies this parallel tank circuit will have a significant increase of voltage due to resonance.

Hi S.G.

Wareagle's "reason" answers the what, but not the why nor the where. The OP asks where does the 'extra' voltage come from. One thing it does not come from is a mathematical representation.

As the OP does not know where the 'extra' voltage comes from, then a digression into complex algebra won't help; the digression necessarily presumes the OP has gotten this far in circuit analysis. And if the OP has gotten this far, then he/she has certainly been exposed to earlier concepts that explain "where it comes from." But having asked this particular question, I'd bet a lot of good money that the OP is no more enlightened by wareagle's 'reason' than he/she was prior to posting the question. Why? Because the 'reason' isn't a reason at all.

It's one thing to sling equations, but if the underlying mechanism is not well understood, the equations represent little more than mathematical obfuscation. I don't believe wareagle has answered the OP's question at all. IMO.

Take care,

-e

Here's a simple answer. AC voltage fluctuates 0 - positive - 0 - negative. The highest voltage across the inductor occurs when the voltage across the resistor is going through 0. So you can't add them together because they occur at different times.

"The highest voltage across the inductor occurs when the voltage across the resistor is going through 0."

All without phasors.

As you know, the voltage across the inductor is higher, equal to, or lower than the voltage across the resistor (and hence the current through it) depending on how fast the voltage across the resistor (and hence the current) passes through 0.

The original poster has clearly measured the voltage across the resistor and across the inductor separately. Measuring the two voltages in this way neglects the relative phase difference between them and because of this - as you correctly pointed out - the sum of the two will not (generally) equal the applied voltage.

The difference between the two voltages depends on the frequency of the applied voltage. At low frequencies (what constitutes "low" being a function of the specific inductance), the voltage across the inductor will be less than the voltage across the resistor.

At sufficiently high frequencies (again, the value of "high" being a function of the inductance), the voltage will be greater.

The two voltages are equal only when the overal resistance equals the reactance (with the parasitic resistance of the inductor's windings included with the component resistance in this case).

Nice, simple explanation, Guest. Here's a GA point for you.

Kind regards,

TV

Dear Guest, this has not yet been said but there is no "extra" voltage. I believe wareagle tried to expalin where you are making the error. All mathematical operations must be performed using phasors because all the quantities have both magnitude and angle. Take the following example:

Suppose you have an RL series circuit with resistance R= 300Ω (300+j0), Inductive reactance X_L= j100 (0+j100) and supply voltage e(t)=20+j0 V (M= 20, Θ = 0) (If this notation does not mean anything to you then you can stop reading and take out the text books again).

The total current in the circuit is i_T= e/Z_T where Z_T = R +j│X_L│= 300+j0+0+j100 = 300+j100 (rectangular form). Converting Z_T into polar form results in M=316.22, Θ=18.43. Therefore the current in the circuit is i_T=20, Θ = 0 / 316.22 , Θ=18.43 => M= 0.063, Θ=-18.43.

The voltage across the resistance is V_R = i_TR = 0.063x 300 Θ=-18.43 +0 => M= 18.9, Θ=-18.43 V (as expected voltage and current are in phase)

The voltage across the inductive reactance is V_L= i_TX_L = 0.063x 100 Θ=-18.43 +90 => M= 6.3,Θ= 71.57. (voltage across the inductor leads the current by 90^o )

Convert V_R and V_L to rectangular form to get V_R = 17.93-j5.97 and V_L= 1.99+j5.97.

We should have e(t)=V_R+V_L.

e(t)= V_R+V_L= 17.93-j5.97+1.99+j5.97 = 20+j0 Ignoring round off error. Therefore the conclusion is that there is no "extra" voltage.

Hopefully this assists you with your query.

Kind Regards

Mr. W.A Snow

Anything given to you by an Auburn fan has to be WRONG by definition! The "extra" voltage is caused by the smoke in the wires. Just don't let it out and everything will be all right...

Where did this very small bit of intelligence come from?

The CIA?

Nice work, Chas.

TV

Thank you all I've been enlightened in so many ways. Thanks again