# Resistors in series vs parallel

I understand that for resistors in series we need to add R for each resistor to get the equivalent resistance, where R is resistance. I also understand the mathematical manipulations that show that for resistors in parallel we need to add the 1/R for each resistor to get the equivalent resistance. However I have a couple of questions that have to do with the physical interpretation of the mathematical manipulations involved.

First, I am trying to understand why in the case of parallel resistors we have to add 1/R's instead of adding R's as in the case of resistors in series i.e. could it be thought that the current in the wire is like a sort of water flow in the pipe that has two horizontal parallel pipes positioned one on top of the other and in between connected by a number of vertical pipes that have a tap each that could be adjusted to represent different resistances. Then at the top of each vertical pipe the water pressure P exists and at the bottom the pressure will be reduced by P*R.

For example, suppose we have three vertical pipes, so then the pressure at the bottom of each (after the tap) is P*R for the total of 3*P*R so why is the equivalent resistance not 3*R*P but 1/R = (1/R1 + 1/R2 + 1/R3) and how can I be able to visualize this relation?

-
As a general addendum to the answers below, I like to think of this as replacing the original $n$ resistors with $n$ identical resistors all having the same "mean resistance". Which mean? The $1$-mean (arithmetic mean) for series and $(-1)$-mean (harmonic mean) for parallel--that can be checked with a bit of algebra easily enough, but recognizing this as just different cases of the generalized mean might slightly alleviate protests of intuition. Another common one in physics is the $2$-mean aka root-mean-square. –  Stan Liou Dec 20 '13 at 3:34
Adding two resisters of same resistance(R) in parallel is equivalent to increase the Cross-section area of a single resister(R) twice. This makes available more carriers hence conductivity is doubled. –  user31782 Dec 20 '13 at 12:35

$$R_{EQ} = R_1 + R_2$$

This follows from KVL and Ohm's Law: $V = IR$. Since series connected circuit elements have identical current $I$ through:

$$V_{EQ} = IR_1 + IR_2 = I(R_1 + R_2) = IR_{EQ}$$

$$G_{EQ} = G_A + G_B$$

This follows from KCL and the dual of Ohm's Law: $I = VG$. Since parallel connected circuit elements have identical voltage $V$ across

$$I_{EQ} = VG_A + VG_B = V(G_A + G_B) = VG_{EQ}$$

Now, it is clear that conductance is the reciprocal of resistance, thus:

$$G_{EQ} = \frac{1}{R_{eq}} = \frac{1}{R_A} + \frac{1}{R_B} \Rightarrow R_{eq} = \dfrac{1}{\frac{1}{R_A} + \frac{1}{R_B}}$$

The physical interpretation is quite straightforward. Adding another path for current allows more current for a given voltage; putting a resistance in parallel is adding a conductance - an extra path for current. This is analogous to adding another path for water flow for a given pressure; allowing more flow for a given pressure.

-

When two resistors are in parallel, the voltage across them is the same, but the current through them might be different. The current through the parallel elements will add to form the total current, giving

$I = \frac{V}{R_1} + \frac{V}{R_2}$.

Manipulating this to get V/I in terms of R1 and R2 will get you the formula for parallel resistors,

$R_{eq} = \frac{V}{I} = \left(1/R_1 + 1/R_2\right)^{-1}$

When two resistors are in parallel, the current through them is the same, but the voltage across them might be different. The voltages add to give the total voltage:

$V = I R_1 + I R_2$

and manipulating this will get us the formula for series resistors,

$R_{eq} = R_1 + R_2$.

-