JEE Main | 2018Current Electricity and Electrical InstrumentHard

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)

Create a free account to view solution

View Solution Free
Topic: Current Electricity and Electrical Instrument·Practice all Current Electricity and Electrical Instrument questions

More Current Electricity and Electrical Instrument Questions

A potentiometer wire of 10 m length and having a resistance of 1 ohm/m is connected to an accumulator of emf 2.2 volt an...In the circuit shown in figure, the 5Ω resistance develops 20.00 cal/s due tothe current flowing through it. The he...A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) hic...A current passes through a wire of non-uniform cross-section. Which of the following quantities are independent of the c...ABCD is a square where each side is a uniform wire of resistance 1Ω. A point E lies on CD such that if a uniform wi...