 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/section-b/8/c/solution

Let

C_{T} = C + C_{1}

be the total parallel capacitance, since capacitors in a parallel circuit have additive capacitances. Put

ω_{*} = 2 π (50 Hz)

C_{T} = \frac{L}{R ^{2} + ω _{*}^{2} L ^{2}} = \frac{0.150}{1 0 ^{2} + [ 2 π ( 50 ) ( 0.150 ) ] ^{2}} F = 64.6368 μ F .

The additional capacitance is therefore

C_{1} = C_{T} - C = 64.6368 - 10.0000 = 54.6368 μ F ≃ 54.6 μ F .

This is the added capacitance, rather than the new total, and the conversion from frequency to angular frequency is essential.

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