JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1

A resistor

R

is connected to a parallel combination of two identical batteries each with emf

E

and an internal resistance

r

. The potential drop across the resistance

R

is:

View answer and explanation

Correct option: A

\frac{2 E R}{2 R+r}

Option A (

\frac{2 E R}{2 R+r}

) is correct because it's derived from calculating the equivalent EMF and internal resistance of the parallel battery combination and then applying Ohm's law.

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