A circuit contains an ideal battery, three resistors and two ideal ammeters. The ammeters read 0.2 amp and 0.3 amp. After two of the resistor are switched (inerchanged) the readings of ammeter do not change. Then

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`.