How is the net resistance lesser than the least resistance when resistances are connected in parallel? The electrons will have to flow through one of the resistances, right? To me, it seems like the net resistance should be equal to the least resistance. Can someone please explain intuitively? (I know how the formula works)
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asked Sep 30, 2015 at 4:19
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2 Answers
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The current flows through both resistors.
What works for me is to use the analogy of fluid through pipes under pressure.
Imagine two huge tanks, connected by two small pipes of different sizes. The pressures in each of the two tanks are analogous to the two voltages. The pipes are analogous to the resistors. The fluid is analogous to the electrons. The flow (mass flux) of the liquid is analogous to the current.
Each of the pipes has its own linear function of flow as a response to pressure, $F_1 = k_1 p\space$ and $F_2 = k_2 p\space$, where the $k$'s are the flow efficiencies of each pipe for letting liquid flow in response to pressure. A larger pipe will be more efficient at letting liquid through than a smaller pipe. The total flow would be the sum of the flows through each pipe. $$F= F_1 + F_2 =k_1 p + k_2 p = (k_1 + k_2) p =k p\tag 1$$ and the total flow efficiency of the two pipe system would be the sum of the two pipe efficiencies, $$k=k_1+k_2\tag{2}$$ Flow efficiency for a pipe is analogous to current conductance (ability to conduct current) for a resistor. It would be great if the convention was to use conductance, but we typically describe components with the attribute "resistance", which is the inverse of conductance.
In this scheme, inverses of the pipe efficiencies, $\frac 1{k_1}$ and $\frac 1{k_2}$, are analogous to the resistances, $R_1$ and $R_2$, of each of the resistors. Conversely, inverse of the resistances, $\frac 1{R_1}$ and $\frac 1{R_2}$, of each of the resistors are analogous to the pipe efficiencies, $k_1$ and $k_2$. Therefore, by analogy to Eq. (2), $$\frac 1 R=\frac 1{R_1}+\frac 1{R_2}\tag 3$$ and by analogy to Eq. (1), $$I= I_1 + I_2 =(\frac 1{R_1}) V + (\frac 1{R_2}) V = (\frac 1{R_1} + \frac 1{R_2}) V =\frac V R\tag 4$$
I know you were not interested in rederiving the formula, put others may want that.
To summarize in words, not formulae: in the plumbing analogy, it is easy to see that the flow between tanks would not just be the flow through the big pipe, but both. And in the circuit, the current would not just be the current through the "easier" resistor, but both. The easier resistor (big pipe) being the one with smaller resistance.
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1$\begingroup$ Thankyou. I could relate to the anology very well. I just want to clarify if I got it right: if a wire splits into two wires, each of resistance r, instead of going through one wire of resistance r, in a single resistor only half the charge needs to flow. Hence the charge is transferred twice as fast. And therefore the net resistance is halved. Right? $\endgroup$ Commented Sep 30, 2015 at 8:44
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$\begingroup$ Halved compared to the resistance of one branch of the split, yes. Is that what you mean? If you mean compared to the resistance of the original unsplit wire, no. If it is a true simple split, the cross-section of the whole wire would be the same as the sum of the cross-sections of the two split piece of wire. Resistance goes as the inverse of the cross-section given a uniform material unless the cross-section is down to a few angstroms, which would be invisibly thin. see next comment $\endgroup$ Commented Sep 30, 2015 at 22:31
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$\begingroup$ 1/R = 1/R' + 1/R" --- Split a wire exactly in half and each half will have twice the resistance of the whole wire: R'=R"=2R. But resistance for the two halves together as a system R* would be R*=1/(1/R'+1/R"). Substitution: R*=1/(1/2R+1/2R)=1/(2/2R)=1/(1/R)=R. IOW, R* for the split system will, as expected, equal R for the original complete wire. If we were doing all this with "efficiencies" as in the plumbing model, it would be more intuitive. But by convention we use resistance, which is the inverse of "efficiency". $\endgroup$ Commented Sep 30, 2015 at 22:49
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$\begingroup$ @user93868, Don't forget to upvote and/or select my answer. $\endgroup$ Commented Sep 30, 2015 at 22:52
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$\begingroup$ @user93868, see added paragraph about "conductivity". $\endgroup$ Commented Sep 30, 2015 at 23:11
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Consider resistors $R_1$,$R_2$,$R_3$ connected in parallel (voltage $V$ is across them, $I$ is the total current from $V$. and $I_1$,$I_2$ and $I_3$ are currents in each resistor) The voltage across each resistor is same, which is $V$. So current would be same through it, even hundreds of resistances are connected in parallel. So if we add more resistances all $I_1$, $I_2$ and $I_3$ would be same, and new current paths are added. So total current would be more. And in a way with the same voltage $V$, we get more current with more resistances in parallel. Hence equivalent resistance is lesser than even the least of them.
