AS Daily A Level Physics question
A uniform metal wire is connected to a power supply with a constant e.m.f. and negligible internal resistance. The wire dissipates power P. The wire is then disconnected and permanently stretched to twice its original length. Its volume is assumed to remain constant. The stretched wire is then reconnected to the same power supply. What is the new power dissipated by the stretched wire?
Answer
The correct answer is A.
The resistance of the wire is given by the formula R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. The volume of the wire, V = L × A, is constant. When the length is doubled (L' = 2L), the cross-sectional area must be halved (A' = A/2) to keep the volume constant. The new resistance R' is therefore R' = ρL'/A' = ρ(2L)/(A/2) = 4(ρL/A) = 4R. The resistance quadruples. The power dissipated is given by P = V²/R, where V is the constant e.m.f. of the power supply. This formula is appropriate because the voltage across the wire is constant. The new power P' will be P' = V²/R' = V²/(4R) = (1/4)(V²/R) = P/4. Therefore, the new power dissipated is one quarter of the original power.