a) A wire of a certain length (a=0.0065/°C) has a resistance of 4Ω (at 20°C). Find its resistance at 80°C.
R=R0(1+aΔT)R=4(1+0.0065×(80−20))R=4×1.39=5.56 ΩR=R_0(1+aΔT) \\ R = 4(1 + 0.0065 \times (80-20)) \\ R = 4 \times 1.39 = 5.56 \; \OmegaR=R0(1+aΔT)R=4(1+0.0065×(80−20))R=4×1.39=5.56Ω
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