three resistors each of resistance 30 ohm are arranged to form an equilateral triangle.a battery of emf 2V and negligible internal resistance is connected between any two vertices of the triangle.what will be the current delivered by the arrangement?

30 ohms and 30 ohms in series:Rse=R1+R2       =30+30       =60 ohms60 ohms and 30 ohms in parallel:Reff=R1R2R1+R2        =60×3060+30        =20 ohmsCurrent:I=VReff =220 =0.1 A

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The arrangement will be like.
Two 30ohm resistors in series and their equivalent in parallel with the third 30ohm resistor.
So your Rnet = 20 ohm.
(Using rules of parallel and series combination )

So Current drawn -

I = V/R

According to question.

I = 2/20
I = 0.1 Ampere.
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