JEE Main 2008PhysicsCurrent ElectricityKirchhoffs Circuit LawsmediumMCQ
JEE Main 2008 — Current Electricity Question with Solution
From: AIEEE 2008
Question
A 5V battery with internal resistance 2Ω and a 2V battery with internal resistance 1Ω are connected to a 10Ω resistor as shown in the figure.
The current in the 10Ω resistor is
This question includes a diagram. The text above accompanies the figure.
Choose an option
▸Show full solutionCorrect option: C
Correct answer
C0.03AP2toP1
Step-by-step explanation
Applying kirchoff's loop law in ABP2P1A we get
−2i+5−10i1=0...(i)
Again applying kirchoffs loop law in P2CDP1P2 we get,
10i1+2−i+i1=0...(ii)
From (i) and (ii)11i1+2−[25−10i1]=0
⇒i1=321 A from P2 to P1
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