 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/section-a/2/solution

Using the convention

e^{iω t}

Z = R + i (ω L - \frac{1}{ω C}), ω = 2 π f .

At the stated frequency,

ω L = 180.956 Ω

and

1/ (ω C) = 88.4194 Ω

. Thus

Im Z = + 92.5 Ω, ∣ Z ∣ = 533 Ω, ar g Z = + 10. 0^{\circ} .

The vector is

(525, 92.5)

in the resistance–reactance plane. Positive reactance means that the net circuit is inductive and current lags voltage. With the alternative

e^{- iω t}

convention the imaginary signs reverse, while the magnitude and physical phase relationship agree.

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