A2 Daily A Level Physics question
In a lab, you charge initially uncharged 22 µF capacitor(s) from a 6.0 V supply through a 100 kΩ resistor by closing a switch. You test three setups using the same resistor and supply: (i) a single capacitor, (ii) two identical capacitors in series, (iii) two identical capacitors in parallel. In each case you record: the initial current I0 just after closing the switch, and the time thalf for the voltage across the left-hand capacitor to reach half of its final steady value. Ignore internal resistance. Which statement must be true?
Answer
The correct answer is D.
Correct: D — I0 is the same in all three, and thalf(parallel) is four times thalf(series). A misunderstanding is that capacitance sets the initial current; at t = 0 the uncharged capacitors act like a short, so I0 = V/R (here 6.0 V/100 kΩ ≈ 60 µA) in all cases; additionally, thalf scales with RC so the parallel:series ratio is 4, not 2. B ignores that thalf to any fixed fraction scales with the time constant: Ceq differs (C/2 in series, 2C in parallel), so the times are not equal. C incorrectly claims I0 depends on total capacitance (it doesn’t at t = 0) and also mis-states the time scaling; in fact thalf(single) ∝ RC, thalf(series) ∝ (1/2)RC, thalf(parallel) ∝ 2RC, giving a 4:1 ratio between parallel and series. D uses the correct limiting behaviour at t = 0 for I0 and the correct proportionality of thalf to RC with Ceq(series) = C/2 and Ceq(parallel) = 2C, hence thalf(parallel)/thalf(series) = (2C)/(C/2) = 4.