how can any circuit be a true parallel circuit?

[mta145]

In a parallel circuit isnt the last load on the circuit always in series making it a series-parallel circuit?

 

[don_resqcapt19]

No.........

 

[quogueelectric]

The wire may be considered in series but that is usually negligable in resistance value.

 

[bob]

In a parallel circuit isnt the last load on the circuit always in series making it a series-parallel circuit?

In series with what. The last load is electrically connected to both conductors that are in parallel. There may be a single conductor from both of the parallel conductors to the load. Is that what you are referring to?

 

[steelersman]

In a parallel circuit isnt the last load on the circuit always in series making it a series-parallel circuit?

No because if it were in series then it would be getting it's supply from the previous loads load.
If they were in series then all loads are connected line to load, line to load, line to load and so forth.

 

[LarryFine]

In a parallel circuit isnt the last load on the circuit always in series making it a series-parallel circuit?

The "last" load will be in series with its switch, if there is one, but that's electrically the same as any other load.

 

[iaov]

If it was in series all the current would be passing through it.

 

[gar]

090221-1753 EST

mta145:

Your last load has wires in series with the load, and their resistance (impedance) determines the voltage across the load. So yes this is a series circuit. Each of the loads before the last also has some wire resistance leading to it and thus each is a series circuit. With many loads your have a big series parallel circuit.

However, from a concept point of view, as Edison understood, if your source impedance is low in comparison to the load impedance, then you consider this as an approximation to a pure parallel circuit. Note: there is no real world pure parallel circuit. In a voltage source power system what is achieved with an approximation to a pure parallel circuit is the ability add and remove loads with little effect on other loads as a result of these changing loads. This is a necessary practical requirement to build a power distribution system.

When you are not concerned about the effect of line voltage drop you approximate the system as a pure parallel circuit.

.

 

[Doug S.]

how can any circuit be a true parallel circuit?

When you are not concerned about the effect of line voltage drop you approximate the system as a pure parallel circuit.

Ever heard of super-conductor(s)(ing)? Otherwise they're all series-parallel. :wink:

Here's what google has to say about it.

 

[mta145]

Thank you gar

Thank you gar

Your explanation was excellent and very helpful to me.

 

[gar]

090221-1936 EST

Doug S.:

I decided to ignore superconductivity because I did not think it would help clairify mta145's question.

.

 

[steelersman]

090221-1753 EST

mta145:

Your last load has wires in series with the load, and their resistance (impedance) determines the voltage across the load. So yes this is a series circuit. Each of the loads before the last also has some wire resistance leading to it and thus each is a series circuit. With many loads your have a big series parallel circuit.

However, from a concept point of view, as Edison understood, if your source impedance is low in comparison to the load impedance, then you consider this as an approximation to a pure parallel circuit. Note: there is no real world pure parallel circuit. In a voltage source power system what is achieved with an approximation to a pure parallel circuit is the ability add and remove loads with little effect on other loads as a result of these changing loads. This is a necessary practical requirement to build a power distribution system.

When you are not concerned about the effect of line voltage drop you approximate the system as a pure parallel circuit.

.

Damn!! I thought I knew what a parallel and a series and a series-parallel circuit were until gar got all up in the mix!! Now I'm officially confused.

 

[glene77is]

The wire may be considered in series but that is usually negligable in resistance value.

Quo,

No, the Load may not be considered in "series".
By "Design", the circuit loads are "parallel".

Think of this 'odd' way, for just a moment:
A three phase motor connected on a Wye transformer
could be seen as connected as three "series" loads.
This does sound 'odd', right?

Comments are welcome.

 

[glene77is]

... until gar got all up in the mix!! ...

Steeler,
Gar has "cognitive flexibility", and a lifetime of experience.
That talent makes his responses interesting.
I find them temporally confusing,
but then I just read again, and further.

 

[gar]

090222-1335 EST

glene77is:

Yes. One transformer secondary is in series with one motor coil. Or if you want the motor coil is in parallel with the secondary. In this case it will be a function of what you want to study and the accuracy of the result.

Consider a simple case: A battery, switch, resistor, and a capacitor all in a series connection. Sometimes this battery will be assumed ideal, usually the case for circuit analysis. Idea means constant voltage independent of load current, and zero internal impedance.

The resistor and capacitor are assumed idea. The resistor is only a resistor and the capacitor is only a capacitor. In real life this is not the case, but may be very close for some components at some frequencies.

Now, let's suppose that all the resistance is the internal resistance inside the battery, and there is no switch. Connect the capacitor to the battery. Relative to the battery terminals we can describe this as a parallel circuit. Seems reasonable.

However, if I want to determine the voltage on the capacitor as a function of time following the connection of the capacitor to the battery, then I would analyze the circuit as a series RC circuit being driven by a voltage source. The analysis will consist of writing an equation like Vbat - iR - q/C = 0. This is the sum of the series voltage drops. This also can be written as R*dq/dt + q/C - Vbat = 0 . For a solution see some books on differential equations. However, by inspection, you can see that the initial current at the time of switch closure with 0 charge on the capacitor is Iinitial = Vbat/R . If the initial charge on the capacitor was such that the capacitor voltage was Vbat/4, then the initial current would be Iinitial = (Vbat - Vbat/4) / R .

Basically you draw an equivalent circuit for what you want to analyze. This is a model of the real world circuit where you include the important components and exclude those you consider unimportant. Then you analyze the equivalent circuit.

For example: A 1000 ft length of coaxial cable may have a capacitance of 50,000 pfd as measured at a low frequency. This capacitance is not concentrated at one small point. Rather it is uniformly distributed evenly over the entire length. At 60 Hz I might assume this is one small capacitor 1" long for my equivalent circuit. However, at 144 mHz I can not do that because the wavelength of this frequency is 2 meters. Now I need to treat this device (coax cable) as a transmission line. If I apply a DC voltage to one end of this cable and a light bulb as a load at the other end, then for steady state purposes I do not care about the capacitance, or transmission line characteristics, but only the total loop resistance.

.

 

[glene77is]

Gar,

Quote""
R*dq/dt + q/C - Vbat = 0
""Quote

Reads like (God's) higher language.
It has been 28 years, but it looks familiar.

Your examples ring true.
Thanks.

 

[Speedskater]

Even a two conductor extension cord can be drawn schematically as a large number of very small resistors, inductors and capacitors.

 

[glene77is]

Speed,
Yes, indeed!
And a calculus description is very handy, also,
since it allows combining the equations in one location,
and calcs the combined effect, to the point.
Amazing way to think!