circuit with a light bulb

Original post by Kallisto

Hello!

I need help in terms of circuits. Here is the task:

A light bulb ( 60 W/ 110 V) is connected to an outled with a voltage U = 220 V (f = 50 Hz). After that an inductor with an inductivity L is connected, thus the light bulb doesn't blow.

a.) Calculate the peak values of the voltages.

U_{0}= U_{eff} * \sqrt{2}

, where

U_{eff}= 110 V

for light bulb and

U_{eff}= 220 V

for the outlet

Is that right?

b.) calculate the Inductivity L and the peak voltage

U_{0L}

and the total voltage.

U_{0L} = \omega * L * I_{0}= 2* \pi * f * L * I_{0}

Is this formula right? If yes, what is

I_{0}

? unfortunately I don't know how to calculate the Inductivity.

Total voltage:

R = R_{1} + R_{2}= \dfrac{U_{1}}{I_{1}} + \dfrac {U_{2}}{I_{2}}

, where

U_{1}

and

Unparseable latex formula:

I_{1}}

are the values of the light bulb and

U_{2}

and

Unparseable latex formula:

I_{2}}

are the values of the outlet. Right?

c.) Now a capacitator is connected parallel to the circuit. Explain why the light bulb may blow.

I guess there are two resistors, namely an inductive and a capacitive one after connection. By this connection the current is looking for the resistor with the small value, the capacitative resistor. If this value of the capacitative resistor is too small, the current is too high, thus the light bulb begins to blow.

Is my consideration right?

d.) Which capacity is required that the light bulb goes out (does not blow)? explain why the light bulb goes out.

As there is an LC circuit, it has to be this formula to calculate the capacity in my opinion:

\omega * L = \dfrac{1}{\omega * C} => C = \dfrac{1}{\omega^2 * L}

is that right?

I guess it has something to do with antiresonance and a trap circuit why the light bulb goes out. Is this consideration right?

Thanks in advance!

Hello.

Your question and response does not use conventional English-grammar or physics/engineering terms which implies a language barrier? I'm guessing you are not a native English speaker or have tried to use a language translator?

You used the word 'inductivity' which I assume to mean inductive reactance?
The same applies to capacitive reactance?

It would help greatly if you could post a scan of the question including the circuit schematic as it is very difficult to understand from the way you described it.

We can then try and work through the answers.

Thanks.

(edited 9 years ago)

Reply 2

OP

Original post by uberteknik

Hello.

Your question and response does not use conventional English-grammar or physics/engineering terms which implies a language barrier? I'm guessing you are not a native English speaker or have tried to use a language translator?

You used the word 'inductivity' which I assume to mean inductive reactance?
The same applies to capacitive reactance?

It would help greatly if you could post a scan of the question including the circuit schematic as it is very difficult to understand from the way you described it.

We can then try and work through the answers.

Thanks.

Yes, you are right. I'm not an English native speaker, so I have had problems with the terms. I have used a dictionary to find the terms out.

No, I meant the inductivity and the capacity, physical values, no more.

Unfortunaly the task doesn't contain a sketch or something like that. As far as I understand a light bulb is connected with an outlet. This is the circuit. And this circuit is connected with an inductor. And the inductivity of this inductor should be calculated. After this calculation a capacitator is added (=connected) to this circuit. And now I should give an explanation why the light bulb can be blow (=destroy).

Hope I was a little bit clearer now.

Reply 3

Original post by Kallisto

Yes, you are right. I'm not an English native speaker, so I have had problems with the terms. I have used a dictionary to find the terms out.

No, I meant the inductivity and the capacity, physical values, no more.

Unfortunaly the task doesn't contain a sketch or something like that. As far as I understand a light bulb is connected with an outlet. This is the circuit. And this circuit is connected with an inductor. And the inductivity of this inductor should be calculated. After this calculation a capacitator is added (=connected) to this circuit. And now I should give an explanation why the light bulb can be blow (=destroy).

Hope I was a little bit clearer now.

OK. Thanks.

Without any further information, we have to make an assumption that the a.c. power source voltage and the lamp ratings are all true r.m.s. values.

a.) Calculate the peak values of the voltages.

U_{0}= U_{eff} * \sqrt{2} , where U_{eff}= 110 V for light bulb and
U_{eff}= 220 V for the outlet

Is that right?

Correct.

To convert the r.m.s. voltages to peak values, multiply by

\sqrt{2}

b.) calculate the Inductivity L and the peak voltage U_{0L} and the total voltage.

U_{0L} = \omega * L * I_{0}= 2* \pi * f * L * I_{0}

Is this formula right? If yes, what is I_{0} ? unfortunately I don't know how to calculate the Inductivity.

Correct.

It's easier if you make some substitutions:

The equivalent d.c. resistance of the inductor at any given frequency is called the Inductive Reactance, X_L

Where

\mathrm{X_{L}} = \omega\mathrm{L} = 2\pi\mathrm{fL}

You can then treat

\mathrm{X_{L}}

as if it were a pure resistance, but only at the same frequency

\mathrm {f}

From ohms law

\mathrm{V} = \mathrm{IR}

and substituting

\mathrm{R} = \mathrm {X_L}

gives

\mathrm{V} = \mathrm{X_{L}I} = \mathrm{2\pi fL I}

what is I_{0} ? unfortunately I don't know how to calculate the Inductivity.

\mathrm{Power} = \mathrm{V I}

then

\mathrm{I} = \frac{V}{P}

(Use the r.m.s. values 110V and 60W for this).

(edited 9 years ago)

Reply 4

OP

Original post by uberteknik

x

I'm back!

Do I have this right, that the inductivity can be calculated by the inductive reactance?

V = R * I

, where R ist the inductive reactance, so

R = 2 * \pi * f * L

V = 2 * \pi * f * L * I => L = \dfrac {V}{2 * \pi * f * L}

Is this right? I wonder what V is (the voltage by any chance?) and how this value can be calculated. What a current is the physical value I exactly?

Was the calculation of total resistance

R = R_{1} + R_{2}

right? I'm so sorry, as I have written total voltage instead of total resistance. My mistake!

I will be back soon and write and ask about the other parts of the circuit!

Reply 5

Original post by Kallisto

I'm back!

Do I have this right, that the inductivity can be calculated by the inductive reactance?

V = R * I

, where R ist the inductive reactance, so

R = 2 * \pi * f * L

V = 2 * \pi * f * L * I => L = \dfrac {V}{2 * \pi * f * L}

Yes, it is correct.

Advice: get into the habit of using appropriate labels such as

X_L

for the inductive reactance, or you will end up making mistakes when doing the calculations.

X_L = 2 \pi f L

(inductive reactance = X_L)

then

V_L = 2 * \pi * f * L * I

(In this case,

V_L

is the r.m.s. voltage potential across the inductor.)

Original post by Kallisto

Is this right? I wonder what V is (the voltage by any chance?) and how this value can be calculated. What a current is the physical value I exactly?

It is best if you draw a phasor diagram and then you will not get confused. The potential voltage developed across the inductor leads the current by 90^o

I =

current through series path and flows through both the inductor and the lamp (Kirchoffs current law)

I_{supply} = I_L = I_{lightbulb}

(all of these are in phase and have the same magnitude.)

The lamp has ratings of 110V and 60W. i.e. the circuit needs to ensure that these are correct for the light bulb to work correctly.

Power = V_{lightbulb}I_{lightbulb}

I_{lightbulb} = \frac{power}{V_{lightbulb}} = \frac{\mathrm{60}}{\mathrm{110}} = \mathrm{0.545 A_{r.m.s}}

then

R_{lightbulb} = \frac{V_{lightbulb}}{I_{lightbulb}} = \frac{\mathrm{110}}{\mathrm{0.545}} = \mathrm{201.67\Omega}

Original post by Kallisto

Was the calculation of total resistance

R = R_{1} + R_{2}

right? I'm so sorry, as I have written total voltage instead of total resistance.

Because the inductor reactance is frequency dependent, then the total circuit resistance is an impedance not resistance.

Z_{total} = \sqrt{X_L^2 + R_{lightbulb} ^2}

We still need to calculate the value of

X_L

and to do that we need to first calculate the phase angle

\phi

between the supply voltage and the p.d developed across the inductor and the light bulb

I'm going to write this out by hand, because Iatex takes too much time and you definitely need diagrams to understand what is going on. I will be back shortly.

(edited 9 years ago)

Reply 6

Original post by Kallisto

x

Here it is:

RL series circuit sht1.jpeg

Reply 7

OP

Original post by uberteknik

Here it is:

RL series circuit sht1.jpeg

Thans ueberteknik! these sketches make the understanding so much easier.

But there are still two tasks left, so back to c.) first.

I don't get it why the light bulb may blow (destroy) when a capacitor is connected to this circut. I'm sure it has something to do with the different reactances and the frequency, but I have no idea why.

Reply 8

Original post by Kallisto

Thans ueberteknik! these sketches make the understanding so much easier.

But there are still two tasks left, so back to c.) first.

I don't get it why the light bulb may blow (destroy) when a capacitor is connected to this circut. I'm sure it has something to do with the different reactances and the frequency, but I have no idea why.

Start thinking about the problem in a different way:

SHORT VERSION

If the supply is suddenly switched off, the inductor no longer sources current from the supply but instead sources current from the capacitor. A back e.m.f. is generated across the inductor which is considerably bigger than the original supply e.m.f. This allows the instantaneous power developed in the light bulb to increase above it's rated power and the light bulb filament could fuse if the charge stored on the capacitor is great enough. (i.e. the bigger the capacitor the more likely the light bulb will blow).

DETAILED DESCRIPTION

The capacitor stores energy (when charged) in the electric field between the capacitor plates.

The average energy stored on the capacitor is

\frac{1}{2}CV_s^2 Joules

.

Since the capacitor is in parallel with the series L-R combination, a conduction path exists between the plates of the capacitor.

Now think about what happens if the original power supply is suddenly interrupted (switched off).

There will still exist a conduction path between the plates of the capacitor via the parallel path through the LR series combination. Current (electrons) are still able to flow around that path to equalise the charge difference between the capacitor plates.

However, the collapse of the power supply voltage will also stop the current flow through the inductor sourced from the power supply. In addition, with the supply e.m.f. removed, there is nothing to support the electric field between the capacitor plates.

The inductor sources charge from the capacitor and will continue to flow because of the p.d. across the capacitor plates. As the charge on the capacitor starts to deplete, the e.m.f. across the inductor begins to fall and with it, the magnetic field tries to fall.

According to Faraday, as the magnetic field falls, a back e.m.f is generated which wants to prevent further collapse of the magnetic field.

E = -\frac{d\Phi_B}{dt} = -L\frac{di}{dt}

. The back e.m.f. rises and so the original current flow at maximum through the inductor tries to be maintained.

Since the charge on the capacitor is continually falling, the back e.m.f. must rise in sympathy in order to maintain the magnetic field. The instantaneous current through the LR path increases rapidly as the inductor tries to prevent the collapse of the magnetic field by sourcing charge from the capacitor, which itself is trying to equalise the charge difference across the capacitor plates in the same direction. i.e. both the capacitor and the inductor are now working for their own ends in mutual support.

The back e.m.f is in exact reverse polarity to the original supply and therefore re-enforces the direction of current required for the discharge between the capacitor plates via the LR path. i.e. the capacitor discharges through the LR path forced by the collapsing magnetic field of the inductor.

The inductor back e.m.f. therefore rises to produce a large voltage spike considerably greater than the original peak supply voltage and in the opposite direction. When the electric field between the capacitor plates reaches zero, the charges on both plates are in equilibrium. However, the magnetic field in the inductor is now at a maximum and that energy is used to continue sourcing charge in the same direction (depletes electrons) from the capacitor which now begins to build charge in the opposite direction. i.e. the stored magnetic field energy is now used to drive charge onto the opposite capacitor plate.

The capacitor E--field in the opposite direction now grows as the inductor B-field collapses, until, the B-field has no more energy at which point the E-field in the capacitor is at maximum but now reversed.

Since

P = \frac{V^2}{R} = I^2R

the instantaneous power developed in the light bulb increases significantly and because of that stored energy available from the capacitor, the discharge can be enough to stress the light bulb filament so that it blows.

N.B. The stored energy in both the capacitor electric field and the inductor magnetic field will now exchange alternately between the two via the circulating charge (current) and since the current path is via the resistance of the light bulb, the energy will dissipate in that resistance as a decaying, exponential envelope, damped oscillation.

If you want the maths for this, I can provide. However it does require you to be very famiiar with the solutions to 2nd order (non-linear)differential equations, Faradays and Lenz's laws for induction, Kirchoffs voltage and current rules.

(edited 9 years ago)

Reply 9

OP

Original post by uberteknik

x

That is a very good and extensive explanation. Thanks! but what the abbreviation e.m.f stands for? what is the meaning?

And now there is just a question left, so back to d.)

First I should calculate which capacity is required so the light bulb goes out (not to fuse).

As there is an electrical impedance between a capacitator and inductor, the capacitatice reactance must be the same the inductive one, so:

X_{C} = X_{L}

, where

X_{C} = \dfrac{1}{\omega*C} = \dfrac{1}{2*\pi*f*C}

and

X_{L} = \omega*L = 2*\pi*f*L

I guess I have to put the calculated value for L in inductive reactance, right?

From this it follows:

C = \dfrac{1}{4*\pi^2*f^2*L}

Is that right?

Next I should explain why the light bulb goes out. Here is my explanation:

There is an electrical impedance with antiresonance, so if the capacity of the capacitatice reactance is great enough, a trap circuit come into being which causes a decrease of the current, thus the current which goes to the light bulb is to weak to make the light bulb shine.

Is this consideration right? is there anything which I have forgotten to mention?

Reply 10

Original post by Kallisto

That is a very good and extensive explanation. Thanks! but what the abbreviation e.m.f stands for? what is the meaning?

E.M.F. = Electro Motive Force. This is the voltage provided by a power source and is a measure of the potential energy in Joules/Coulomb of charge carried by electrons around a closed circuit.

Original post by Kallisto

And now there is just a question left, so back to d.)

First I should calculate which capacity is required so the light bulb goes out (not to fuse).

I think there is a translation problem with the wording.

As long as the power source provides current, the light bulb will NOT go out.

The only way for the light bulb to turn off is for the power supply to be interrupted.

Original post by Kallisto

As there is an electrical impedance between a capacitator and inductor, the capacitatice reactance must be the same the inductive one, so:

X_{C} = X_{L}

, where

X_{C} = \dfrac{1}{\omega*C} = \dfrac{1}{2*\pi*f*C}

and

X_{L} = \omega*L = 2*\pi*f*L

I guess I have to put the calculated value for L in inductive reactance, right?

From this it follows:

C = \dfrac{1}{4*\pi^2*f^2*L}

Is that right?

If the circuit comprised perfect lossless inductance and capacitance only, the equation is correct.

However, there is the problem of the light bulb resistance to consider.

X_C is calculated for a very specific reason: to reduce the phase angle between the voltage and current in the LR series combination such that power is consumed by the light bulb resistance only and nowhere else.

When the correct capacitor is chosen, the phase angle between the voltage and current across the L and C reactances are reduced to zero (or as close as possible to zero). i.e. the phase angles generated by the reactive components (between the voltage and currents of those reactances) cancel out.

The capacitor cancels the effect of the inductor by creating a resonant condition as explained in my description previously. Energy is exchanged between the capacitor and inductor with a common current path via the lamp resistance.

If the inductor and capacitor are both perfect devices (i.e. no energy loss) then the only power used in the circuit would be in the lamp resistance.

Once again, the lamp will continue to shine with 60W only taken form the supply and all of that power used in the lamp resistance and nowhere else in the circuit.

As I said, the lamp will turn off ONLY when the power supply current is interrupted (switched off) and at no other time.

As I also previously stated, turning off the power supply will induce a large back e.m.f across the inductor causing an increase of current through the light bulb resistance. In this way, the energy stored in the capacitor will be dumped in the light bulb resistance in a short span of time. It is this instantaneous power which ruptures the light bulb filament causing the light bulb to be destroyed.

I will be back shortly with how to calculate the correct value of C.

(edited 9 years ago)

Reply 11

OP

Original post by uberteknik

x

To sum up your explanation in short and from the view of the capacitator reactance only:

The turning off the light bulb depends on the 'right' capacitor reactance, because if the capacitator reactance has the 'right' value of the capacity, the phase angle between the voltage and the current across the inductive and capacitatice reactance are reduced to zero, as the phase angle is generated by these reactive components. So the light bulb turns off. Am I right?

And know I will calculate the value of C:

If I put the the calculated value of L in the equation of C, I get:

C = \dfrac {1}{4*(\pi*50 Hz)^2*1.11 H} = 9.13 * 10^-6

Is that the right value of C? did I have understand your explanation well?

Reply 12

Original post by Kallisto

To sum up your explanation in short and from the view of the capacitator reactance only:

The turning off the light bulb depends on the 'right' capacitor reactance,

Not really.

When the capacitor reactance at 50Hz is correctly matched with the inductor-light bulb impedance at 50Hz, then the only power consumed by the circuit will be that of the light bulb. (60W from the supply of 220V.) i.e. the 110V light bulb works correctly even though the supply is 220V and still at the correct 60W output rating. No other power is used in the circuit.

Original post by Kallisto

because if the capacitator reactance has the 'right' value of the capacity, the phase angle between the voltage and the current across the inductive and capacitatice reactance are reduced to zero, as the phase angle is generated by these reactive components.

NO. The phase angles across the reactive components are not reduced to zero. (Current will always lag the voltage across the inductor and always leads the voltage for the capacitor).

As a result of the inductive and capacitive reactance, the voltage across the resistor caused by their reactances sum to produce a time varying voltage across the resistor. Also, because the current and voltages due to the reactance have a phase relationship, they will also modify the phase relationship between the voltage and current in the resistor.

It's the power developed in the resistor as a result of those different phase angles caused by the reactances which cancel and it is the reactive power which is reduced to zero, nothing else. This applies to the first paragraph above only at a fixed frequency of 50Hz.

Original post by Kallisto

So the light bulb turns off. Am I right?

NO.

Go back to the light bulb in series with the inductor only. The inductor reactance is frequency dependent and governed by:

X_L = 2\pi fL

As the frequency and/or the inductance increases, then so does the reactance.

The inductor in series with the light bulb resistance forms a frequency dependent potential divider. If the inductor reactance increases, then more of the supply voltage will be dropped across the inductor and less across the light bulb. i.e. the power developed by the light bulb resistance (P = V²/R must reduce and so the light bulb will dim.

i.e. if the frequency of the supply is kept constant at 50Hz, then increasing the value of L > 1.1H will cause the light bulb to dim and eventually (when L is great enough), the light bulb will first glow red and as L continues to increase, will stop producing light altogether.

Original post by Kallisto

And know I will calculate the value of C:

If I put the the calculated value of L in the equation of C, I get:

C = \dfrac {1}{4*(\pi*50 Hz)^2*1.11 H} = 9.13 * 10^-6

Is that the right value of C?

The value you calculated above is for a resonance condition in a circuit with a perfect inductor in parallel with the capacitor and no other resistance (like that of the light bulb).

i.e. X_L = X_C only. Under these conditions, when the supply is removed, current will continue to circulate alternately between the capacitor and inductor with no losses. i.e. oscillation will continue indefinitely.

However, the circuit DOES have a resistance: the light bulb resistance which must be taken into account.

For maximum efficiency, the capacitor reactance X_C is chosen so that the power factor of the inductor-light bulb combination (true power/apparent power = 60/120 = 0.5) is brought as close as is possible to unity (60W true power/60W apparent power). i.e. the efficiency of the circuit with the inductor-light bulb alone (50%) is brought to 100% by introducing the parallel capacitor.

X_C = \frac{E^2}{reactive power Q} = \frac{E^2}{I^2X_L} = \frac{(220)^2}{(0.545)^2 349.3} = 466.5\Omega

But

X_C = \frac{1}{2\pi fC}

C = \frac{1}{2\pi f X_C} = \frac{1}{2\pi 50*466.5} = 6.83\mu F

This means that the 110V light bulb will produce an output of 60W from the 220V supply, with a total circuit power consumption of 60W when using a series inductor of 1.1H and a parallel capacitor of 6.8 microFarads as long as the supply frequency is maintained at 50Hz. (i.e. close to 100% efficiency with no losses)

(edited 9 years ago)

Reply 13

OP