IIT JEE , AIEEE Forum - Physics , Electricity

v_gurucharan

Q.13. Two cells E1 and E2 in the given circuit diagram have an emf of 5 V and 9 V and internal resistance of 0.3 and 1.2 ohmrespectively. Calculate the value of current flowing through the resistance of 3 ohm.

v_gurucharan

i m desc the figure:

E1 and E2 are connected in opposite polarity wrt each other in series.
the resistance of 6ohm and that of 3 ohm are connected in parallel to each other but in series with the batteries and resistance of 4.5 ohms.

tarinbansal

EDITED---------

tarinbansal

If the two cells are connected in series, there net emf can be found out be simple addition/subtraction of their individual emfs.
As they r connected in opposite polarities, the net emf= 9-5=4V.
Put a new battery of 4V in place of two with the same polarity as that of 9 V battery.,
Add their internal resistances and put that net resistance(0.3+1.2=1.5) in series and solve the circuit.

Aatish

I hope that this is only the circuit you are talking about.....

so...Net EMF, E = 9 - 5 = 4volts
then....net internal resistance r = 1.2 + 0.3 = 1.5 ohm
and......net resistance other than r, R = Rp + 4.5 = 6.5 ohm

so....net voltage across the circuit V = E.R / (R+r)
so V = 3.5 volts

and thus current through 3ohm resistance is,

i = 3.5 / 3 = 1.167 amperes.......

I hope this is correct.......

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