The resistance of the meter bridge AB in given figure is 4 `Omega` .With a cell of emf `epsilon=0.5 V` and rheostat resistance `R_(h) = 2 Omega ` the null point is obatined at some point J is found for `R_(h) = 6 Omega " The emf "epsilon _(2)` is
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C
At pull point , current `i=6/((R_h+4))` For balance : `epsilon=i` Where x is the resistance between A and I. So, Case 1 : `0.5=(6x)/(2+4) rArr ` x=0.5 Case 2: `epsilon_2=(6x)/10 rArr epsilon_2` =0.3 volts .
Find the current drawn from a cell of emf 2 V and internal resistance 2 Omega connected to the network given below.
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VMC MODULES ENGLISH-JEE Main Revision Test-6 | JEE-2020 -PHYSICS
