Steven W. answered • 07/15/16
Tutor
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(4,315)
Physics Ph.D., college instructor (calc- and algebra-based)
I take it the lamps are 10 Ω each.
When n resistors are combined in parallel, the expression for the equivalent resistance of all the resistors together is:
(1/Req) = (1/R1) + (1/R2) + ... +(1/Rn)
This has the effect, by the way, of assuring the equivalent resistance of all the resistors in parallel is less than the resistance of any of the individual resistors alone. This is in contrast to putting resistors in series, where the equivalent resistance is the sum of the individual resistances, and is thus greater than any of the individual resistors alone.
So if you want to effectively cut the strength of a group of resistors, connect them in parallel. To increase their strength, connect them in series.
For parallel connection, given the equivalent resistance expression above, a useful shorthand for n identical resistors connected in parallel is that:
Rew = R/n
So, for having 1 through 6 resistors in parallel, just take the 10 Ω for a single resistor and divide it by the number in parallel to get he equivalent resistance in each case.
Once you have converted your parallel combinations in to one equivalent resistor, you now have, in effect, a circuit with a single battery and a single resistor with that equivalent resistance. You can solve for the current (I) coming out of the battery in this case by just using:
Vbat = IReq
Because we did not do involve the battery at all, or do anything to it, to calculate the equivalent resistance, that current coming out of the battery calculated using Req is the same current coming out of the battery in the original circuit, with all the resistors in parallel. And it is thus the total current in the circuit in that original case.
Hope this help! If you have more questions about this, or would like to compare answers, just let me know.
Steven W.
tutor
If you understand the principles, then that is most of the battle. The ,ath can be tricky in parallel because of all the inverses, which can make the algebra seem more mysterious when trying to find the solution.
To work out the math, let's start with an example that is not exactly the same as your problem: a 12 V battery powering two resistors of 4 Ω and 12 Ω connected in parallel.
First, what is the equivalent resistance of the two resistors?
By rule: (1/Req) = (1/R1) + (1/R2), so we have (1/Req) = (1/4) + (1/12)
Depending nn the numbers, at this point, you could just enter the right-hand numbers directly into your calculator and crunch our a solution (I have students who do this, and it is fine). But, with these numbers, we can also proceed "analytically," by finding a common denominator and using that to combine the terms on the right
(1/Req) = (1/4)+(1/12) --> (1/Rew) = (3/12) + (1/12) = 4/12
Thus, (1/Req) = (4/12), which means Req = 12/4 =
3 Ω
Then, figuring the current just means using Vbat = IReq, for the reason I described above. So 12 V = I(3 Ω), so I =
4 A.
If you added more resistors in parallel, you would just add more terms on the right above before either computing the right side with your calculator or determining the common denominator.
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Now, if the resistors are equal, calculating the equivalent resistance of two of them in parallel gives a special, simplifying case.
(1/Req) = (1/R) + (1/R)
In this case, the right side already has a common denominator, so you can just add them directly.
(1/Req) = (2/R), and thus Req = (R/2).
With three equal resistors in parallel, this becomes:
(1/Rew) = (1/R) + (1/R) + (1/R) = (3/R), and thus Req = (R/3)
This is why, in general, for n equal resistors in parallel, the equivalent resistance is Req = R/n.
If you put in your values for R and n for each case, that should give you the answers you listed. Then using Vbat = IReq allows you to calculate the current.
Note the general pattern where the equivalent resistance goes down as you add more in parallel, and thus the current drawn from the battery goes up.
Just let me know if you have more questions!
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07/16/16
Bobby A.
07/15/16