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Two cells e(1) and e(2) connected in opp...

Two cells `e_(1) and e_(2)` connected in opposition to each other as shown in figure. The cell of emf 9 V and internal resistance `3Omega` the cell is of emf 7V and internal resistance `7Omega`. The potential difference between the points A and B is

A

8.4V

B

5.6V

C

7.8V

D

6.6V

Text Solution

Verified by Experts

The correct Answer is:
A
`I=(Deltaepsi)/(r_(1)+r_(2))=(9-7)/(3+7)=(2)/(10)=0.2A`
Potential difference across cell `epsi_(1)` is `=9-0.2xx3=9-0.6=8.4V`
Potential difference across `epsi_(2)`,
`V_(AB")=epsi_(2)+0.2 r_(2)=7+0.2xx7=7+1,4=8.4V`
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Knowledge Check

  • In the circuit shown in fig. the cell has emf 10V and internal resistance 1 Omega

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