Dorsa R.
asked • 05/30/20Two identical batteries, each with an emf of 18 V and an internal resistance of 1 , are
wired in parallel by connecting their positive terminals together and connecting their negative
terminals together. The combination is then wired across a 4-ΩΩ resistor. The current in the 4-Ω
resistor is:
Dr Gulshan S. answered • 05/31/20
PhD In Physics and experience of teaching IB Physics and Math High sc
The total effective resistance of two cell in parallel = R1*R2/ (R1 + R2)
= 1/2 = 0.5
The net EMF in parallel = Value of either of batteries = 18V
Thus total resistance of circuit = 4 Ohm
So
I = V/ Total resistance
I = 18/ 4.5 = 180/45 = 4 amp
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Yuri D.
This is a wrong solution, but you obtained the correct answer(s). You replaced the real circuit by an equivalent circuit {"two ideal EMF's, connected in parallel" -- in series with "two the two internal resistances connected in parallel" -- in series with "external resistance"}. This is not the correct replacement. In reality the equivalent circuit must be: { [ "ideal EMF1 in series with internal resistance1" connected in parallel with "ideal EMF2 in series with internal resistance2" ] -- in series with "external resistance R".} You can then use Kirchhoff's rules 1 (junction rule) and 2 (voltage rule) to one of the single loops with the external resistor. And indeed the correct answer for the current through the external resistor is 4 A .05/07/22