# Parallel circuits - Overall resistance decreases with additional resistor

Let's say that there is a parallel circuit with two identical resistors in parallel with each other. If a third resistor, identical to the other two, is added in parallel with the first two, the overall resistance decreases.

Why does this overall resistance decrease?

-
– Qmechanic♦ Mar 10 at 18:27

1
This question is straight out of chapter 19 of the third edition of Matter & Interactions by Chabay and Sherwood. It is associated with figure 19.48. Since it's a direct plea for a homework question answer, I think this question should be closed. There is at least one other one for which I similarly recommend the same. – user11266 Mar 10 at 22:51
– Chris Mar 11 at 0:27
Please take a look at other questions that have been closed for this very reason. – user11266 Mar 11 at 1:06

## 5 Answers

For resistors $R_1, R_2, \dots, R_N$ in parallel, the equivalent resistance $R_e$ is given by $$\frac{1}{R_e} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_N}$$ If two resistors with equal resistance $R_1 = R_2 = R$ are in parallel, then this gives $$R_e^{(2)} = \frac{R_1R_2}{R_1+R_2} = \frac{R^2}{2R} = \frac{R}{2}$$ If three identical resistors $R_1 = R_2 = R_3 = R$ are in parallel, then the equivalent resistance is $$R_e^{(3)} = \frac{R_1R_2R_3}{R_1R_2 + R_2R_3 + R_3R_1} = \frac{R^3}{3R^2} = \frac{R}{3}$$ In fact, for $N$ identical resistors one has $$\frac{1}{R_3^{(N)}} = \frac{N}{R}$$ so that $$R_e^{(N)} = \frac{R}{N}$$ and therefore the resistance decreases with the addition of each successive resistor in parallel.

-
This answer doesn't answer the question that was asked; it merely restates the question in mathematical terms and provided no mechanism. Besides, it's a homework question from Matter & Interactions. – user11266 Mar 10 at 22:00

3
@JoeH The question was "Why does this overall resistance decrease?" which, in my opinion, is sufficiently vague that it affords the following interpretation, which is the one I took: "Given that we know how to mathematically determine the resistance of parallel resistors, how do we show that adding a resistor to a set of resistors in parallel will decrease its resistance?" I respect that you may have interpreted the question differently, and I'm glad you added a distinct response. Given that I received the check, perhaps my interpretation was closer to the intent of the author? – joshphysics Mar 10 at 22:27
I didn't think the question was vague at all. The OP is asking questions straight out of the M&I textbook and there is a unique story line which the authors follow. If the instructor is implementing M&I correctly, then any answer that includes resistance won't be acceptable. Look for other questions by this same poster. This particular question is from chapter 19 and is associated with figure 19.48. – user11266 Mar 10 at 22:50

2
@JoeH I think you make fair points. I did, in fact, restate the question in mathematical language, and I also did answer the question having restated it in that form. I feel that both my response, and yours, are useful. Cheers^2! – joshphysics Mar 10 at 23:08

1
YAY! We agree on something! :) – user11266 Mar 10 at 23:10
show 3 more comments

Why does this overall resistance decrease?

A more elegant, sophisticated way to see why is through the notion of duality.

In electric circuit theory, conductance (the reciprocal of resistance) is dual to resistance. Other dual pairs are:

voltage - current

series - parallel

inductance - capacitance

Thevenin - Norton

and so on ...

For example, consider Ohm's Law: $v = iR$. The dual is: $i = vG$

You probably intuitively understand that adding a resistor in series increases the total resistance.

The dual of this is adding a conductance in parallel increases the total conductance.

But, if the conductance increases, the reciprocal, i.e., the resistance, decreases.

Mathematically:

Conductances in parallel add:

$G_{total}=G_1 + G_2 + G_3 = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} = \dfrac{1}{R_{total}}$

or

$R_{total} = \dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}$

Now, it's clear that adding another resistor in parallel, increases the denominator thus decreasing the total resistance.

-

The reason adding resistors in parallel decreases overall resistance is that adding resistors in parallel increases the effective cross sectional area of the circuit. Current is proportional to cross sectional area, so the overall current drawn from the battery (assuming a DC circuit) increases. However, adding a resistor in parallel doesn't change the potential difference across the new resistor or the other existing parallel resistors (that follows from the very definition of parallel). If the circuit obeys the macroscopic Ohm's law ($\Delta V = IR$), then resistance is the quotient of potential difference and current. Potential difference doesn't change when adding a new parallel resistor, but current increases so the quotient decreases. The accepted answer is mathematically correct, but doesn't actually address your question.

-

Suppose we have a simple circuit with only one resistor. Assume the voltage across the circuit is 10V, and current is say, 2Amps. Now assume we add an additional component to the circuit in parallel as in the following diagram:

The voltage accross each component is not reduced by this action, and now we have a 10V pull over both components in the circuit.

If we consider the first component, we can use the formula $V = I_1R_1$, to find the current to be $I_1 = \frac{V}{R_1}$ (This will equal 2Amps, juas as in the original circuit). Similarly, we can find the current over the second component using the formula $V=I_2R_2$ and therefore $I_2 = \frac{V}{R_2}$. The voltage here is the same as before, because adding additional resistors in parallel does not reduce the pulling power of the battery.

The total current in the circuit is then obtained by adding the current flowing in each of the components, to give $I_T = I_1 + I_2 = \frac{V}{R_1} + \frac{V}{R_2} = V(\frac{1}{R_1} + \frac{1}{R_2})$.

Now we can think of the two resistors as one big resistor, of resistance $R_T$, and use the formula $V = I_TR_T$.

Therefore, $R_T = \frac{V}{V(\frac{1}{R_1} + \frac{1}{R_2})} = \frac{1}{(\frac{1}{R_1} + \frac{1}{R_2})}$, which is the formula used in one of the above answers.

We now can see intuitively why the resistance has decreased. This is because adding the second resistor allowed for an additional current $I_2$, which combines with the original current, $I_1$, to form a larger current than before. When current increases over a constant voltage, we can say that the total resistance of the circuit has decreased.

-

I almost didn't answer, but maybe it can help someone better understand the accepted answer.

Some people will cringe at the thought of this, but I often find myself comparing electricity to plumbing. It works reasonably well in your question and helps me visualize joshphysics' correct answer.

It is the same as a large tub filled with holes. Adding more drain pipes will cause the tub to drain faster. In this case, the drain pipes are more resistors. Adding a tiny drain hole (representing an additional resistor of large value) will undeniably result in more current flow.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9492347836494446, "perplexity_flag": "middle"}