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Two cells having an internal resistance ...

Two cells having an internal resistance of `0.2 Omega` and `0.4 Omega` are connected in parallel, the voltage across the battery is 1.5 V. If the emf of one cell is 1.2 V, then the emf of second cell is

A

2.7 volt

B

2.1 volt

C

3 volt

D

4.2 volt

Text Solution

Verified by Experts

The correct Answer is:
B

`E_(eq) = (E_(1)/r_(1) + E_(2)/r_(2))/(1/r_(1) + 1/r_(2))`
`rArr 1.5 = (r_(2)E_(1) + E_(2)r_(1))/(r_(1) + r_(2))`
`rArr 1.5 = (0.4 xx 1.2 + E_(2) xx 0.2)/0.6 rArr 0.9 = 0.48 + E_(2) xx 0.2`
`rArr E_(2) = (0.9 - 0.48)/0.2 = 2.1` Volt
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