JEE Main | 2018Current Electricity and Electrical InstrumentEasy

Question

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

Options

A.

11.5 V and 11.6 V

B.

11.4 V and 11.5 V

C.

11.7 V and 11.8 V

D.

11.6 V and 11.7 V

Solution

121+13211+12=373volt

req = 2×12+1=23Ω

Now its equivalent circuit is :
i=37/310+23=3732
∴ V10Ω=i×10=3732×10=37032 = 11.56 volt
Hence (1)

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