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Two batteries with e.m.f 12 V and 13 V a...

Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of `10 Omega`. The internal resistances of the two batteries are `1 Omega` and `2 Omega` respectively. The voltage across the load lies between :

A

11.6 V and 11.7 V

B

11.5 V and 11.6 V

C

11.4 V and 11.5 V

D

11.7 V and 11.8 V

Text Solution

Verified by Experts

For parallel combination of cells,
`E_(eq)=(E_1/r_1+E_2/r_2)/(1/r_1+1/r_2) therefore E_(eq)=(12/1+13/2)/(1/2+1/2)=37/3 V`
Potential drop across `10Omega` resistance
`V=(E/(R_(eq))) times 10=(37//3)/((10+2/3)) times 10=11.56V`
Alternative method
Applying KVL in loop ABCFA,`-12+10(I_1+I_2) +1 times I_1=0`
`implies 12=11I_1+10I_2` .......(i)
Similarly in loop ABDEA, `-13+10(I_1+I_2)+2 times I_2=0`
`implies 12=10I_1+12I_2`..........(ii)
SOlving Eqs (i) and (ii) we get `I_1=7/16A,I_2=23/32A`
`therefore` VOltage drop across `10Omega` resistance is `V=10(7/16+23/32)=11.56V`


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