Uncovering the Physics Behind 1/x + 1/x = 1/y

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In summary: Summary: In electronics, resistors and inductors in parallel and capacitors in series are calculated using the sum of the reciprocals formula. Similarly, with lenses, the focal point is calculated using the reciprocal sum of the distances to the image and object. This formula appears in different places due to the simplicity of the A=BC law, where keeping B constant results in a sum law and keeping A constant results in a reciprocal sum law. This concept can also be applied to other laws, such as V=IR and Q=VC. In parallel circuits, the effective resistance is calculated by adding the conductances, while in series, the resistances are added. The same concept applies to lenses, where the diopters are added
  • #1
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In electronics the addition of resistors and inductors in parallel (and of capacitors in series) has the formula of the sum of the reciprocals of the resistance equals the reciprocal of the total.

With lenses the formula for the focal point is again the reciprocal of the sum of the reciprocals of the distances to the image and to the object.

Is there some underlying physics principle behind this formula that allows it to show up in different places? I got that in general the total is inversely related to the number of terms, but is there a stronger connection here?


tl;dr Why does this formula show up in different places, 1/x + 1/x = 1/y ?

thank you for any insight.
 
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  • #2


When you put resistors in parallel and wish to calculate the resistance of the circuit you are really interested in the conducting ability.
You therefore calculate the conductance of each resistor by taking its inverse, add these together to get the total conductance then take the inverse to express this as resistance.
 
  • #3
hi antistrophy! :smile:

it's because the simplest formula is A = BC, and if you keep B constant, you get a sum law, but if you keep A constant you get a reciprocal sum law …

for example with resistors we have the law V = IR, so …
in series, I is constant, so it's sums

in parallel, V is constant, so it's reciprocal sums​

(similarly for inductors, using dI/dt instead of I)

but with capacitors (through which there's no current) we have the law Q = VC, so …
in parallel, V is constant, so it's sums

in series, Q is constant, so it's reciprocal sums​

A = BC laws are fairly common … that's why sum and reciprocal sum laws keep appearing! :biggrin:

(I can't think off-hand what the A = BC law for lenses is :redface:)
 
  • #4


AH HA! 100% clear as day, thank you very much, well done. I suspect the law for the lenses has something to do with the magnification being equal to the image distance over the object distance. Thanks again tiny tim.
And thanks Syhprum for your reply as well.
 
  • #5


tiny-tim said:
hi antistrophy! :smile:

it's because the simplest formula is A = BC, and if you keep B constant, you get a sum law, but if you keep A constant you get a reciprocal sum law …

for example with resistors we have the law V = IR, so …
in series, I is constant, so it's sums

in parallel, V is constant, so it's reciprocal sums​

(similarly for inductors, using dI/dt instead of I)

but with capacitors (through which there's no current) we have the law Q = VC, so …
in parallel, V is constant, so it's sums

in series, Q is constant, so it's reciprocal sums​

A = BC laws are fairly common … that's why sum and reciprocal sum laws keep appearing! :biggrin:

(I can't think off-hand what the A = BC law for lenses is :redface:)

hi tim, i don't really get it

i udnerstand if V=RI
then if I is constant in series, that means
Vtotal = IR1 + IR2 + ... since I is constant

but what about parallel

where V is constant?

how does the effective resistance become 1/R = 1/R1+R2+...
 
  • #6


It's because the sum going on is a sum over the currents-- the total I = I1+I2+..., and since with constant V, each I1 = 1/R1 (let's say V=1 to get rid of it), whereas the total I=1/R, where R is the effective resistance.

Another way to say this is, when you have a series circuits, each resistor adds more total resistance, but when you have a parallel circuit, each resistor adds another path for the current to take, so you are really adding conductances not resistances. You always get an inverse sum when the additive parameter is actually inverse to the parameters you are expressing things in terms of, so it's a combination of what syhprum and tiny-tim said.
 
  • #7
to put it in symbols …

if K is conductance (I'm avoiding "C" since I don't want it to be confused with capacitance),

then V = IR becomes I = VK,

and you can clearly see that I = VK with V constant will behave the same way as V = IR with I constant :smile:
 
  • #8


ah i see... haha... i feel that the more i learn the more complicated i see things and simple things are harder to visualize :(
 
  • #9


That suggests that the simple things that used to be easy to visualize were not being visualized correctly. Often, things seem simpler when we understand them, but if they seem too simple, perhaps our understanding is illusory!
 
  • #10


In the case of two objects the second one can be considered as helping the first one. The objects have parameters like ohms or focal length. In the case of the resistance in a series circuit they simply add resistances to get a total.
Put them in parallel and the second one is helping the first one to conduct current, but the only number printed on the outside is resistance.
You must take the reciprocal of each one to get its conductance and you can add those directly. But the answer now is in mho's, (1/r) so you have to take the reciprocal again.
In the case of a lens, the easy measure of it is the focal length which is a measure of the weakness of the lens (it takes so much longer to bring the light to a focus). So when you combine two lenses the focal length is of no use. Here again you take the inverse of the focal length of each one, add those together, and take the reciprocal of the answer. The optics business has a way of dealing with this: they divide the focal length into 1 meter and call it diopters. A 5 diopter lens has a focal length of 8 inches. You can add diopters directly for two lenses in combination.
So if the object has the wrong designation you have to use reciprocals.
 
  • #11


jackpol said:
In the case of two objects the second one can be considered as helping the first one. The objects have parameters like ohms or focal length. In the case of the resistance in a series circuit they simply add resistances to get a total.
Put them in parallel and the second one is helping the first one to conduct current, but the only number printed on the outside is resistance.
You must take the reciprocal of each one to get its conductance and you can add those directly. But the answer now is in mho's, (1/r) so you have to take the reciprocal again.
In the case of a lens, the easy measure of it is the focal length which is a measure of the weakness of the lens (it takes so much longer to bring the light to a focus). So when you combine two lenses the focal length is of no use. Here again you take the inverse of the focal length of each one, add those together, and take the reciprocal of the answer. The optics business has a way of dealing with this: they divide the focal length into 1 meter and call it diopters. A 5 diopter lens has a focal length of 8 inches. You can add diopters directly for two lenses in combination.
So if the object has the wrong designation you have to use reciprocals.

ah i see thanks!
 

FAQ: Uncovering the Physics Behind 1/x + 1/x = 1/y

What is the significance of the equation 1/x + 1/x = 1/y in physics?

The equation 1/x + 1/x = 1/y has a wide range of applications in physics, specifically in circuits and wave propagation. It helps in understanding the behavior of electrical circuits and the properties of waves such as sound and light.

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The equation 1/x + 1/x = 1/y is closely related to Ohm's Law, which states that the current through a conductor is directly proportional to the voltage and inversely proportional to the resistance. In this equation, x represents resistance and y represents current, making it a fundamental part of Ohm's Law.

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