# A quiz that I give all my electronics students at mid-term

#### KL7AJ72

Joined Apr 15, 2021
22

You have a 12volt 25 watt lamp that you need to run off of house current (120VAC, 60 Hz in the U.S.). You can drop the voltage to the proper voltage efficiently using just series reactance. With the circuit above, what value of L1 do you need to make the lamp happy? (Capacitor C1 is non-negotiable.)

#### KL7AJ72

Joined Apr 15, 2021
22
View attachment 301458
You have a 12volt 25 watt lamp that you need to run off of house current (120VAC, 60 Hz in the U.S.). You can drop the voltage to the proper voltage efficiently using just series reactance. With the circuit above, what value of L1 do you need to make the lamp happy? (Capacitor C1 is non-negotiable.)
P.S.: Show your work!

#### crutschow

Joined Mar 14, 2008
33,363
Sorry but solving impractical circuit problems just for the sake is solving them has never been my cup of tea.
(I hated them when I was in school).

For the circuit above, all you need is a capacitor of the correct size.
Nobody in real life would likely add an expensive and large inductor for that purpose.

But if someone else wants to solve the problem, go right ahead.

#### Ian0

Joined Aug 7, 2020
8,947
You wouldn't be happy if you touched point #2.

#### Pyrex

Joined Feb 16, 2022
214
Hi,
this is a quiz only. For students...
Ok, lamp current is 55W/12V= 4.58A
Lamp resistance is R=12V/4.58A=2.6R
Capacitor reactance is Xc=1/(2*pi*C)=1/(2*3.14*60*10*10-6)=265R
Inductor reactance is Xl=2*pi*L
Current of the circuit is 120V/Z= 120V/(R^2+( l Xl-Xc l )^2)
Xl=239R, L=0.63H
Take in mind , there is the module in the formula X= Xl- Xc. Z=sqrt( R^2 + X^2 )

#### Ian0

Joined Aug 7, 2020
8,947
Hi,
this is a quiz only. For students...
Ok, lamp current is 55W/12V= 4.58A
It's a 25W lamp!

#### Pyrex

Joined Feb 16, 2022
214
It's a 25W lamp!
Sorry, it a Monday
Anyone can do the job for a 25W lamp

#### Sensacell

Joined Jun 19, 2012
3,330
Sorry but solving impractical circuit problems just for the sake is solving them has never been my cup of tea.
(I hated them when I was in school).

For the circuit above, all you need is a capacitor of the correct size.
Nobody in real life would likely add an expensive and large inductor for that purpose.

But if someone else wants to solve the problem, go right ahead.

Yeah, somehow this seems like a disservice to students, because this solution is totally impractical in the real world.
At least issue a disclaimer: "this idea is not appropriate for any real, practical situation" - so they don't embarrass themselves on their first real job.

#### Ian0

Joined Aug 7, 2020
8,947
You need |Z|=57.6Ω, and Xc = -265jΩ, so Xl has to be 207jΩ. However Xl is likely to be quite a high tolerance and will vary with current. A 10% variation in XL will vary Ztotal between 37jΩ and 78jΩ which would represent a variation of about ±50% in lamp current (and you will need a 1000V rated 10uF capacitor)

#### Ian0

Joined Aug 7, 2020
8,947

#### KL7AJ72

Joined Apr 15, 2021
22
Your Monday is our public holiday!
So far, everyone has missed the point that this has TWO solutions.

#### Ian0

Joined Aug 7, 2020
8,947
So far, everyone has missed the point that this has TWO solutions.
Yes, Last Monday in May and Last Monday in August.
Oh - you mean the circuit.

Two solutions - one either side of resonance,
That's why I put |Z|=57.6Ω, because Z can be ±57.56jΩ
Either solution is equally impractical.
And, the lamp might be happier if it wasn't illuminated - certainly I wouldn't like to be that hot!

#### Alec_t

Joined Sep 17, 2013
14,009
@KL7AJ72
As the lamp type is unspecified, let's assume it's an old-school fluorescent one with a ballast inductor. How does that affect the answers?

#### Jerry-Hat-Trick

Joined Aug 31, 2022
451
In addition to being impractical messing with mains voltage is not clever. I’m not impressed.

#### Ya’akov

Joined Jan 27, 2019
8,550
I doin’t really want to jump on the cynicism train, but for students in particular this seems a very bad idea. It is not only a literally useless example for useful math, it is a dangerous thing. It suggests that this might be a good idea, and of course it isn’t.

Math is already an abstraction, using examples that don’t help tie it to practical application makes it even harder to connect with the real world. So on two grounds—safety and pedagogy—this seems a failure.

#### Ian0

Joined Aug 7, 2020
8,947
120V mains is relatively benign compared to the voltage in the resonant circuit!

#### Ya’akov

Joined Jan 27, 2019
8,550
120V mains is relatively benign compared to the voltage in the resonant circuit!
"I connected things directly to the 120V mains when I was a kid and it didn't do anything bad to me".

But, there were at least a couple of close calls, and few explosive events. While you are certainly correct, it is also important to note that even 120V can be deadly. Much worse is 2KV from an oscilloscope power supply that I only needed one mistake in handling to never make that mistake again—and fortunately, not because I couldn't.

#### Tonyr1084

Joined Sep 24, 2015
7,560
A quiz that I give all my electronics students at mid-term
I've never seen a lamp symbol like that before. Are you really an instructor?
Is this homework that YOUR instructor gave you and you need to find a solution?

#### Janis59

Joined Aug 21, 2017
1,783
But what is the question???
If question is - calculate the amperage, then it is simple form 11 physics task must to be every secondary school pupil be able to solve if want be able to not fail the State Exam!
Just calculate X=-X(C) +X(L), then apply Pytagorus trousers Z=sqrt(X^2+R^2) and apply Ohm`s law i=EDS/Z, or if the question was about separate components then real component i(R)=EDS/R and reactive component i(X)=EDS/X.

And resonance - oh shame the resonance! Resonance is the State where X(C)==X(L) and ANYTHING where it not true is NOT the resonance will You say its near or far of it. Simply resonance here is heavily MISGUIDING term.
However in that serial semi-resonant or even far off the resonant circuit for sure will happen the Voltage Multiplication, thus the 24 Volt input may ocassionally turn to generate type of 24 kV Voltage if very misluck.

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#### Alec_t

Joined Sep 17, 2013
14,009