1plz reply my quesn frn ds its a previous iit quesn
find d emf[V] & internal resistance [r] of a single battery which is equilant to a parallel combination of 2 batteries of emf's v1 &v2 & internal resistances r1 &r2 resply with polarities as shown in fig
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9 Answers
1r1 n r2 in d fig is to show d internal resistance????????????
1since d neg terminal is connected to pos terminal of other battery
jus add d two ptential to get d total emf
i.e
v=v1+v2.....
solvin d rest wait....
no resistors in d circuit?????
1srry i ve posted d last in terms of series connection... ignore dat post...
1luk, ill make it more simple for u...
when we have E1,E2,E3...En and r1,r2,r3,...rn,
and ri not equal to 0, the equivalent battery is given by,
[E1/r1 +E2/r2....En/rn]*{1/[1/r1 +1/r2 +1/r3.....+1/rn]}
taking the polarity signs into concideration...ie if the polarity was opposite we use -E instead of E taking a direction positve by default,
here all we have to do is,
{E1/r1-E2/r2}*[r1r2/(r1+r2)]
=(E1r2-E2r1)/(r1+r2)
and plz dont call me bhai...it gives me a impression as tho' i am sum tapori or sumthing...(jus kidding)
1and the internal resistanse of such an eq. battery is the parallel restance of all the individual resistors...ie.
r1r2/(r1+r2)