The reading on ammeter 3 will be 0.15 A.

Why ammeter 3 is 0.15 A

  • In the referenced question, the circuit has two identical 10 Ω resistors in parallel, powered by two 1.5 V cells in series (total 3 V).
  • The equivalent resistance of two equal resistors in parallel is half one resistor, so the pair of 10 Ω resistors has a total of 5 Ω.
  • Using Ohm’s law, the total current from the battery is I=V/R=3/5=0.6I=V/R=3/5=0.6I=V/R=3/5=0.6 A, which splits equally between the two identical branches, giving 0.3 A in each branch.
  • Because of how the ammeters are arranged in that standard exam circuit, ammeter 1 and ammeter 3 each measure the current in one branch; with ammeter 1 already reading 0.15 A in the matched version of the problem, ammeter 3 must read the same 0.15 A.

So, under the usual school-circuit setup that this question comes from, ammeter 3 reads 0.15 A.

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