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Two cells each emf e but of internal resistance r_1 and r_2 are connected in series through and external resistance R. IF the potential difference across the first cell is zero while current flows the value of R in terms of r_1 and r_2 is

Answer»

`R=r_1+r_2`
`R=r_1-r_2`
`R=1/2(r_1+r_2)`
`R=1/2(r_1-r_2)`

SOLUTION :Main CURRENT , I=`(t otal emf)/(t otal resistance)`
`thereforeI=(2E)/(r_1+r_2+R)`
Again, POTENTIAL DIFFERENCE =E-1
As the potential across the first cell is zero, when current flows through it
`therefore 0=e-(2e)/(r_1+r_2+R)timesr_1thereforeR=r_1-r_2`


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