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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Physics
Chapter
Current Electricity

JEE Advanced 2011 · Paper 2 · Physics · Current Electricity

Two batteries of different emfs and different internal resistances...

Two batteries of different emfs and different internal resistances are connected as shown. The voltage across ABA B in volt is Two batteries of different emfs and different internal resistances are connected as shown. The voltage across A B in

Correct answer

5

Step-by-step solution

VAB=V_{A B}= Equivalent emf of two batteries in parallel. =E1/r1+E1/r21/r1+1/r2=(6/1)+(3/2)(1/1)+(1/2)=5 V\begin{aligned}& =\frac{E_1 / r_1+E_1 / r_2}{1 / r_1+1 / r_2}\\ & =\frac{(6 / 1)+(3 / 2)}{(1 / 1)+(1 / 2)}=5 \mathrm{~V} \end{aligned}\thereforeAnsweris5.AnalysisofQuestion(i)InmyopinionthisisthesimplestquestionofJEE2011.(ii)Whennocurrentisdrawnfromthisequivalentbattery,thenVAB=V=EOtherwise,Answer is 5 . Analysis of Question (i) In my opinion this is the simplest question of JEE 2011. (ii) When no current is drawn from this equivalent battery, then V_{A B}=V=E Otherwise,V=E \pm$ ir (iii) Similar type of question was asked in JEE 1981 and cancelled paper of JEE 1997.
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