AS Daily A Level Physics question
In a lab, a 9.0 V DC bench supply is connected to a single 18 Ω resistor R1; an ammeter in series with R1 reads the current in R1. Without changing the supply, a second identical 18 Ω resistor R2 is connected in parallel across the same supply, while the ammeter remains measuring only the current through R1. Leads and internal resistance are negligible. Which statement must be true for the current in R1 after R2 is added?
Answer
The correct answer is D.
Correct: D — New:old current in R1 = 1:1 (unchanged). In parallel the potential difference across R1 stays at 9.0 V, so I_R1 = V/R = 9.0/18 = 0.50 A both before and after; only the total supply current changes (to 1.0 A). A confuses splitting of total current with the current in one unchanged branch: adding a parallel path adds another 0.50 A branch but does not halve the original 0.50 A. B assumes the reduced total resistance forces more current through each branch; however the branch pd is unchanged so the current in R1 does not double. C mixes series and parallel effects; a 3:2 change would require a change in pd across R1, which does not occur in this ideal parallel addition. D is correct because the voltage across each parallel branch equals the supply, so the current in the original branch remains V/R.