Choice D is correct because:
- The problem requires us to determine which circuit arrangement will result in the largest current reading on the ammeter, given that the ammeter has a resistance of 2 Ω, and the same potential difference is applied across points P and Q in each circuit. We need to calculate the total resistance of each circuit and then use Ohm's Law (I=RV) to find the current through the ammeter. Since the potential difference V is the same for all circuits, the largest current will occur when the total resistance is the smallest.
- Circuit A: The 1 Ω and 2 Ω resistors are in series with the 2 Ω ammeter. The total resistance is RA=1+2+2=5Ω. The current through the ammeter is IA=5V.
- Circuit B: The 1 Ω and 2 Ω resistors are in parallel. The equivalent resistance of the parallel combination is Rparallel=1+21×2=32Ω. This equivalent resistance is in series with the 2 Ω ammeter. The total resistance is RB=32+2=32+36=38Ω. The current through the ammeter is IB=38V=83V.
- Circuit C: The 2 Ω resistor is in parallel with the 2 Ω ammeter. The equivalent resistance of this parallel combination is Rparallel=2+22×2=44=1Ω. This equivalent resistance is in series with the 1 Ω resistor. The total resistance is RC=1+1=2Ω. The current through the ammeter is IC=2V. However, the current IC splits between the 2 Ω resistor and the 2 Ω ammeter. Since they have equal resistance, the current divides equally. Therefore, the current through the ammeter is IAmmeter=2IC=2×2V=4V.
- Circuit D: The 1 Ω and 2 Ω resistors are in parallel. The equivalent resistance of the parallel combination is Rparallel=1+21×2=32Ω. This equivalent resistance is in parallel with the 2 Ω ammeter. The total equivalent resistance is RD=32+232×2=32+3634=3834=84=21Ω. The total current ID=21V=2V. The current through the ammeter can be found using current division: IAmmeter=ID×Rparallel+RAmmeterRparallel=2V×32+232=2V×3832=2V×82=2V×41=2V.
- Comparing the currents through the ammeters:
- - IA=5V
  - IB=83V=103.75V
  - IC=4V=102.5V
  - ID=2V=105V
- Since 2V is the largest current, Circuit D gives the largest reading on the ammeter.
Choice A is incorrect because:
- In circuit A, the resistors and the ammeter are in series. The total resistance is higher than in circuit D, resulting in a lower current reading on the ammeter.
Choice B is incorrect because:
- In circuit B, the resistors are in parallel, and this combination is in series with the ammeter. The total resistance is higher than in circuit D, resulting in a lower current reading on the ammeter.
Choice C is incorrect because:
- In circuit C, one resistor is in parallel with the ammeter, and this combination is in series with the other resistor. While the total resistance is lower than in circuits A and B, the current through the ammeter is less than in circuit D due to current division.

Electric Current | 2010 - M/J - 9702/11 | Q36

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