# Stupid Series/Parallel Question

Discussion in 'The Projects Forum' started by bakgerman, Aug 19, 2014.

Aug 19, 2014
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2. ### alfacliff Well-Known Member

Dec 13, 2013
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because 2 24 volt things in series will not work on 24 volts. the motor takes 24 volts, and the solenoid switch takes 24 volts. put them in series and they take 48 volts. also the current is not the same for either of them.

3. ### bakgerman Thread Starter New Member

Aug 19, 2014
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Hi alfacliff,

Thanks for your quick reply. Please let me understand. Both the pump and solenoid valve require 24V.

when I put them in series 24V goes in to the first component (soleoid valve) does 24 V not come out? What happens to the current?

Likewise in the parallel what happens.

Thank you so much.

Oct 2, 2009
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5. ### bakgerman Thread Starter New Member

Aug 19, 2014
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OK I am getting it.

Both of my components require 24V. I can only achieve this if they are parallel so the Voltage will not split. They will work.

Last question. What happens to the current? I know 1.5A goes into the node before the parallel split. I read on wiki i1 - i2 - i3 = 0. I do not know what is the R for both of my components. I know one needs 0.3A and the other 0.7A. so does the current split evenly 1.5A/2 or does the one that requires more current magically get more current?

6. ### MrChips Moderator

Oct 2, 2009
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No. No. No.

Current, voltage and resistance are governed by Ohm's Law.

I = V/R

Your solenoid requires 0.3A @ 24V
The pump requires 0.7A @ 24V

Together they need 1A @ 24V

The power suppy will provide the 1A @ 24V

Your power supply is designed to put out 24V, 0-1.5A, depending on how much is demanded. If only 1A is needed, it will supply only 1A.

7. ### MrChips Moderator

Oct 2, 2009
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Still doesn't work like that. You cannot put the two in series because they demand different currents.

8. ### bakgerman Thread Starter New Member

Aug 19, 2014
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I hated my electrical engineering classes.

9. ### MrChips Moderator

Oct 2, 2009
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Don't be so hard on yourself. Stick around here long enough and some of it will rub off.

Since you are into hydraulics, the concepts are much the same.

10. ### bakgerman Thread Starter New Member

Aug 19, 2014
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Thanks Mr. Chips.

The main reason it works is because Voltage does not split in parallel. But in series V is consumed by the first component and not enough left for the second component. Is that correct?

11. ### MrChips Moderator

Oct 2, 2009
12,648
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Not quite.

Think of a power supply as a circulating pump that can only provide a certain amount of water pressure. The water (current) flows in a continuous loop so each coil gets the same current flow. The voltage (pressure) doesn't get consumed. When you put two loads in series, they together reduce the current flow. Twice the load means half the current. The pressure is shared across the two loads. If the loads were identical, the current flow would be reduced to 50% and the pressure would be shared equally by the two loads.

In your case, when you put the two loads in series, the current drops to 0.2A which is not enough to power either of the two loads. That is, 0.2A is lower than both 0.3A and 0.7A. So neither solenoid or pump works.

12. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
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No, In the parallel circuit, both loads have 24V across the two terminals. In the series circuit, the sum of the two voltages adds up to 24V

Stare at this simple LTSpice simulation. Note that I represent the two loads by their resistance (R=E/I). R1 = R3 = 80Ω (the solenoid). R2 = R4 = 34.28Ω (the pump).

In the series circuit, the voltage across R1 + the voltage across R2 = 24V, or to say this another way, the voltage across R1 = 24-V(b) = 16.801V. The voltage across the two loads divides inversely proportional to their respective resistance...

In the series circuit, the current through R1 = current through R2, or I(R1) = I(R2) = 0.21001A

In the parallel circuit, I(R3) = V(c)/R3 = 24/80 = 0.3A
I(R4) = V(c)/R4 = 24/34.28 = 0.7A

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13. ### crutschow Expert

Mar 14, 2008
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Basically. But voltage isn't "consumed" by the component, it just appears as a drop across the component's resistance. And the order of the component in series (whether it's first or second) makes no difference. Each will drop voltage proportional to their respective resistance. Thus, in series, if one has twice the resistance of the other then it will drop twice the voltage of the other, and the sum of the two voltage drops will equal the voltage supply.

A simple analogy is to think of electricity as water flowing in a pipe under pressure. If it goes through two valves in parallel, then the pressure (voltage) across each valve is the same but the flow (current) is proportional to the valve orifice size (resistance). If the valves are in series, then the flow through both is the same but the pressure drop across each is proportional to the valve orifice size.