Parallel Equivalent

Thread Starter

Manu1

Joined Mar 30, 2016
23
Basic circuit understanding. If you can't find a parallel equivalent value of two resistors, there is a whole slew of parameters that you won't be able to find.
So you are saying about this Parallel Equivalent of 2k 2k is equivalent to 1k of series ? Sorry I am very Beginner
 

ian field

Joined Oct 27, 2012
6,539
So you are saying about this Parallel Equivalent of 2k 2k is equivalent to 1k of series ? Sorry I am very Beginner
Maths is not my strong point - so I cheat.

Think of an arbitrary voltage and use Ohm's law to calculate the current for each value of resistance, add the sum of all those currents and use Ohm's law to convert the result back to resistance.

When all the resistances are equal - even I can do it without cheating.

If you parallel 2 equal resistors, you get half the resistance. With 3 its 1/3..........and so on.

Its worth looking online for the formulae chart, its often presented as a circle with 4 quadrants, and includes power equations too. There's one pasted to the side of my PC because its easier to just look at it than try to remember.

A basic aid to memory is; "E over IR". If you write that on paper, its easy to see how to transpose the equation for each of the 3 variables.

It doesn't include power, but all you need to remember is; VA=W and use the same rules to transpose for any of the other values.

You end up handling squares and square-roots - but its not so bad once you get the hang of it.
 

alfacliff

Joined Dec 13, 2013
2,458
two identical resistors in paralel equal one half the resistance of one. 2 2k in paraless equals 1k. \.

2 1 ohm in paralell equals 1/2 ohm. sometimes resistors are paralelled to increase power dissapation.
 

SLK001

Joined Nov 29, 2011
1,543
So you are saying about this Parallel Equivalent of 2k 2k is equivalent to 1k of series ? Sorry I am very Beginner
Yes, that is exactly is what I am saying. One equation that you should commit to memory, is E=I⋅R.

Okay, first, let's set the voltage to 2 volts. Now, compute the current in the left 2k resistor. It is 2V=I⋅2KΩ, or I=2V/2kΩ, or, I=1mA. Now compute the current in the right 2k resistor. Again, it is 1mA. Now sum the currents at the bottom node. Here you have 1mA + 1mA, or a 2mA sum. Since the sum of the current going out of a network is the same as the amount going in, we know that the voltage supply is supplying 2mA total. To find the equivalent resistance that would also cause the voltage supply to put out 2mA is also found using the formula above, E=I⋅R. Plugging in our values, 2V=2mA⋅R, or R=2V/2mA, or R=1000Ω.
 

ian field

Joined Oct 27, 2012
6,539
Yes, that is exactly is what I am saying. One equation that you should commit to memory, is E=I⋅R.

Okay, first, let's set the voltage to 2 volts. Now, compute the current in the left 2k resistor. It is 2V=I⋅2KΩ, or I=2V/2kΩ, or, I=1mA. Now compute the current in the right 2k resistor. Again, it is 1mA. Now sum the currents at the bottom node. Here you have 1mA + 1mA, or a 2mA sum. Since the sum of the current going out of a network is the same as the amount going in, we know that the voltage supply is supplying 2mA total. To find the equivalent resistance that would also cause the voltage supply to put out 2mA is also found using the formula above, E=I⋅R. Plugging in our values, 2V=2mA⋅R, or R=2V/2mA, or R=1000Ω.
That's pretty much what I said.
 

crutschow

Joined Mar 14, 2008
24,405
You can also determine the parallel value by calculating the current through each resistor.
Then you add the currents, using basic algebra, to calculate the equivalent single resistance.

For example, for 1k and 2k resistors in parallel:

Itotal = (V/1k + V/2k) = (2V/2k + V/2k) = 3V/2k = V/(2/3k).

Thus the parallel equivalent is 2/3k, the same a calculated from the formula: Rp = (1k * 2k)/(1k+2k) = 2/3k
 

ian field

Joined Oct 27, 2012
6,539
You can also determine the parallel value by calculating the current through each resistor.
Then you add the currents, using basic algebra, to calculate the equivalent single resistance.

For example, for 1k and 2k resistors in parallel:

Itotal = (V/1k + V/2k) = (2V/2k + V/2k) = 3V/2k = V/(2/3k).

Thus the parallel equivalent is 2/3k, the same a calculated from the formula: Rp = (1k * 2k)/(1k+2k) = 2/3k
Form an orderly queue please..........................
 

Thread Starter

Manu1

Joined Mar 30, 2016
23
Okay first thanks all for help this dummy.. So series resistor gives the same value but in parallel the resistor draws haf of the current each right?
 

Thread Starter

Manu1

Joined Mar 30, 2016
23
Okey guys I am very bad at math how ever I am trying to learn it. The question is simple is parallel circuit resistors draw haf of the amount each and is equivalent and why? Is it because it has low current supply?
 

GopherT

Joined Nov 23, 2012
8,012
Okey guys I am very bad at math how ever I am trying to learn it. The question is simple is parallel circuit resistors draw haf of the amount each and is equivalent and why? Is it because it has low current supply?

Think if them as conductors instead of resistors.

So, make a simple circuit with a 2v battery and a (2k ohm) conductor that allows 0.001 amps of current to flow,

Add a second 2k ohm conductor in parallel, that one will also allow 0.001 amps to flow.

Now, calculate the equivalent resistance if you still have 2V battery and a total of 0.002 amps flowing.
 

ErnieM

Joined Apr 24, 2011
8,020
Okey guys I am very bad at math how ever I am trying to learn it. The question is simple is parallel circuit resistors draw haf of the amount each and is equivalent and why? Is it because it has low current supply?
Has nothing to do with there being a low current supply. Actually your diagram shows a battery and a battery can supply a huge current, at least for a little while.

Each of the 2K resistors draw some current. The term "parallel equivalent" asks the question "what single resistor will act the same (draw the same current) as the two resistors?"

Here you have a simple case with two of the same value. The equivalent resistor must draw twice the current as a single resistor, so it needs to be half the resistance.

Keep at trying to understand ohms law: it is quite useful, in fact the single most useful thing you need to learn electronics.
 
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