Institutt for fysikk, NTNU TFY4155/FY1003 Elektrisitet og magnetisme Vår 2007

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Kirchhoff’s voltage rule (K2) yields

E = \frac{Q}{C} = RI_{R}

whereas Kirchhoff’s current rule (K1) yields

I = I_{C} + I_{R}

Furthermore, we have

I_{C} = \frac{dQ}{dt}

Hence:

I_{R}(t) = \frac{V_{0}}{R} cos ωt
Q(t) = V_{0}C cos ωt
I_{C}(t) = −ωCV_{0} sin ωt = ωCV_{0} cos(ωt + π/2)

Total current delivered by the voltage source is then

I(t) = \frac{V_{0}}{R} cos ωt − ωCV_{0} sin ωt

We want I(t) on the form

I(t) = I_{0} cos(ωt − α)

with amplitude I_{0} = V_{0}/Z, where Z is the impedance of R and C coupled in parallel, whereas α becomes the phase angle, i.e., the phase difference between E(t) and I(t). We have

cos(ωt − α) = cos ωt cos α + sin ωt sin α

Hence, by direct comparison:

\frac{cos α}{Z} = \frac{1}{R}
\frac{sin α}{Z} = −ωC

These two equations, with two unknowns Z and α, are easily solved, and we find

Z = \frac{R}{\sqrt{1 + (ωRC)^{2}}}

I_{0} = \frac{V_{0}}{R} \sqrt{1 + (ωRC)^{2}}
α = − arctan(ωRC)

In the limit ω → 0 we should recover well known results from the DC examples given in the lectures, and so we do: Z → R and α → 0 so that I_{0} → V_{0}/R. All the current runs through the resistor R, while the capacitor C now represents an open circuit, where no direct current can run.

With given numerical values, we have

ωRC = 2π · 10^{6} · 10 · 16 · 10^{−9} = 1.0

so that

I_{0} = \frac{1.0}{10} · \sqrt{2} = 0.14 A
α = − arctan 1.0 = −45^{∘}

Sketch of E(t), I(t), I_{R}(t) and I_{C}(t):

Sketch of α, I_{0} and Z (with α in radians and ωRC between 0.016 and 160 along the horizontal axis):

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