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A cell E(1) of emf6 Vand internal resist...

A cell `E_(1)` of emf6 Vand internal resistance `2 Omega` is connected in opposition, as shown in figure, with a cell `E_(2)` of emf 4 V and internal resistance `8Omega` . The potential difference across points A and B is

A

`2.0 V`

B

`5.6V`

C

`10.0 V`

D

`3.6 V`

Text Solution

Verified by Experts

The correct Answer is:
B
Since the cells are connected in opposition, effective emf `=6-4=2V`

`:.` Current `l=(2)/(r_(2)+r_(2))=(2)/(2+8)`
`rArr l=(2)/(10)=0.2A`
Terminal voltate of `E_(1)` is `V_(1)`
`V_(1)=E_(1)-lr_(1)`
`V_(1)=6-0.2xx2`
`=6-0.4=5.6V`
Terminal voltage of
`V_(2)=E_(2)+lr_(2)=4+0.2xx8`
`rArrV_(2)=4+16=5.6V`
:.` Potential difference between A and B= 5.6 V
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