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November 29, 2014

November 29, 2014

Posted by **Sally** on Thursday, March 24, 2011 at 2:13pm.

Three resistors connected in parallel have individual values of 4.0, 6.0 and 10.0 W, respectively. If this combination is connected in series with a 12.0 V battery and a 2.0 W resistor, what is the current in each of the resistors and what is the total resistance of the circuit?

a. 0.59 A

b. 1.0 A

c. 11.2 A

d. 16.0 A

- Physics-Please help -
**Sally**, Thursday, March 24, 2011 at 2:15pmPlease irnore the multiple choice answers-they were for a different problem that I copied down

- Physics-Please help -
**bobpursley**, Thursday, March 24, 2011 at 2:57pmfind total resistance of the parallel network. I assume you mean ohms, not watts on the reistors.

1/Rt=1/4+1/6+1/10=1/60 (15+10+6)=31/60

Rt=60/31

Now add that inseries to 2 ohm, or total resistance is 151/60

current= V/R=12(60/151) check that

now, having total current, how does it divide in the paralel branch.

the voltage across the parallel branch is Rt*It= 60/131*69/151)

so current in each resistor is that voltage divided by either 4,6, or 10 ohms

- Physics-Please help -
**Sally**, Thursday, March 24, 2011 at 3:07pmThank you for your help-I can really use the explanation for my other problems too.

Thank you again-

- Physics-BobPursley please recheck -
**Sally**, Thursday, March 24, 2011 at 3:27pmI have another question-I looked closely at the explanation and when you added the three resistors together you came up with Rt=60/31. Then you added the series to the 2Ohm and said total equals 151/60-I'm lost there because 2 Ohms would equal 30/60 or when added to the inseries it would become resistance total = 60/61 so where does the 151/60 come from? Sorry if I'm just not understanding

- Physics-You don't have to respond-I figured it all out -
**Sally**, Thursday, March 24, 2011 at 6:12pmI figured it out-thank you

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