A circuit contains an ideal battery , three resistors, and tow ideal ammeters. The ammeters read 0.2 A and 0.3 A. After two of the resistors are switched, the readings of the ammeters did not change. Find the battery current.

Options

A. If `R_2 = R_1, I_1 = I_2 = 0.15 A` and `I_3 = 0.05A`.
B. If `R_2 = R_1`, the battery current is `I = 0.20 A` .
C. If `R_2 = R_3, I_2 = I_3 = 0.1 A` and `I_1 = 0.2A`.
D. `If `R_2 = R_3`, the battery current is `I = 0.4A`.

Answer

A, C, D

Explanation

a., c., d.
As the ammeters are ideal, we can conclude that the three
resistors `R_1, R_2, and R` are actually connected in parallel, the
equivalent resistance being
`R_(eq) = (R_1R_2R_3)/(R_1R_2+R_1R_3 + R_2R_3)`
In the figure, the electric currents on each resistors and battery
are shown. The readings of the two ammeters are different and,
consequently, by simple symmetry considerations, resistors
`R_1 and R_3` are necessarily different. In other words, if we
interchange `R_1 and R_3`, the readings of the ammeters will also
be interchanged. Therefore, we have two possible alternatives.
`R_2= R_1` (if resistors `R_2 and R_1` switched, the readings of the
ammeters do not vary) or `R_2 = R_3 (R_2` and `R_3` can be switched
and the ammeters do not change.) The two solutions are :
If `R_2 = R_1` we obtain `I_2 = I_3 = 0.1A` and `I_1 = 0.2A`, and the
battery current is `I = 0.4A`.