IIT JEE , AIEEE Forum - Physics , Electricity

priyesh

Find the net E.M.F of the three batteries shown in the figure

The resistances are internal

Click on the figure to enlarge

tarun007

dude the figure is not clear.......

it is not clear dat which battery has positive term on right n which one has the same on left

priyesh

hey tarun click on the figure & then open it to enlarge

catch_arnnie

solution::

for the internal resistances in parallel, the equivalent resistance is 1/2.

now, according to the formula, E_eq/r_eq = E₁/r₁ + E₂/r₂ where E represents emf & r represents internal resistance.

2E_eq = 2/1 + (-6)/1

E_eq = -2 V

now, the cell on left has emf = 4V

=> net E of circuit = 4 + (-2) = 2 V (ans.)

plz correct me if i'm wrong....

tarun007

hey dude..........solved it

connect the above system wid a 1 ohm resistor.......

solve the circuit using kirchoff rules....... get the amount of current flowing through the the 1 ohm resistor in ampere........dat is numerically equal to the emf of the above sys.

m gettin the emf as 2V..........
and the left side is on higher potential

do rate me if it was helpful

phyana

its a simple application of krichofs rule.

let A,A1,A2 be the current through the cell 4V,2V,6V

then using three loops u get the equations.its vrey diifcult to show my calculation as it may not agree to ur loops.

u get values for A1,A2,A3

find v1=a2*1ohm
the final answer is v1+4v

plz try it

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