EK Physics Lecture 7, 7th ed, page 131, question 159...

[computerdorkdan]

The question says you're adding an additional resistor in parallel then asks what happens to the circuit.

The answer says that "adding a resistor in parallel decreases the overall resistance, which increases current and power". Is that right? Doesn't adding an additional resistor increase overall resistance and decreases current.

Thanks...

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[loveoforganic]

Rule for adding resistors in parallel:

Total resistance = 1 / [(1 / R1) + (1 / R2) + .....)]

So say you have a resistor of 100 ohms, and you add a 200 ohm resistor in parallel.

1 / [(1 / 100) + (1 / 200)] = 66.67 ohms, less than either resistor alone.

You can think of the current as water flowing through pipes - you're adding an extra pipe the water can flow through.

 

[physics junkie]

Firstly, you can rearrange the equivalent resistance formula posted above to the following:

Req = R1*R2/(R1+R2)
It's just a simpler way of looking at the same formula where Req means "equivalent resistance".

Now let's do a bit of factoring and pull the R2 out of the denominator.

Req = R1*R2 / R2*(R1/R2 + 1)
If you divide out the R2's you see that the original formula is equivalent to:
Req = R1 / (1+R1/R2)

(Feel free to test any/all of these formulas to see that they give you the equivalent results for the 100ohm/200ohm example posted above)

With this last formula it is obvious to see that by adding a resistor in parallel you decrease your equivalent resistance via multiplying your previous resistance with one resistor(R1) by a factor of 1/(1+R1/R2)...since you're dividing by a number greater than 1 your equivalent resistance gets smaller.

Hope this helps.

PS How high can you count in base 16? Am I missing the joke?