A2 Daily A Level Physics question
A camera flash circuit charges a capacitor of capacitance C through a resistor R from a DC supply to its final voltage. To reduce shock risk, the designer plans to halve the supply voltage while doubling the capacitance (R is unchanged). The flash tube discharges much faster than the charging time. Compared with the original design, which statement must be true about (i) the energy stored at full charge and (ii) the time to reach 50% of the new final voltage?
Answer
The correct answer is C.
Correct: C — Energy halves; the time to 50% of final voltage doubles. Halving the voltage and doubling the capacitance gives an energy ratio of 2 × (1/2)^2 = 1/2, and the time to reach any fixed fraction of the final voltage scales with RC, so with C doubled (R unchanged) it doubles. A … Wrong because energy scales with C and the square of voltage (not CV), so it would not be unchanged, and the time to a given fraction depends on RC, not on the absolute target voltage, so it would not halve. B … Wrong because it ignores the doubled C: halving V alone would quarter the energy, but doubling C makes the energy half; the time is not independent of C. C … Correct for both the energy change (1/2) and the fraction-of-final-voltage time scaling (factor 2 with doubled C). D … Wrong because while the energy halving is correct, the time to a given fraction is not unchanged; it increases in proportion to C, so it doubles.