Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 5C (Differential Equations)

With no electrical resistance or applied electric potential difference, divide the equation by the inductance $L$ to get a simple harmonic oscillator with angular frequency ${\omega }_0=(LC{)}^{-1/2}$. Its general real solution isEach nonzero solution has period $2\pi \sqrt{LC}$. Here ${\omega }_0$ is an angular frequency; the ordinary frequency in cycles per unit time is ${\omega }_0/(2\pi )$.

The PDF gives $V=H(t)$, the Heaviside step function; the TeX incorrectly changes this to $V=I(t)$. A unit voltage step has derivative the Dirac delta function, so the current is the causal Green function for $L{d}^2/d{t}^2+Rd/dt+1/C$.

The current must be continuous at zero: a jump would create a delta derivative through $L{I}^{\prime \prime }$, absent on the right-hand side. Integrating across zero therefore gives $L[I^{\prime }]+R[I]=1$, so $I(0^{+})=0$ and $I^{\prime }(0^{+})=1/L$. PutThese distinct negative characteristic roots give the overdamped RLC responseTogether with $I(t)=0$ for $t<0$, this solves the step response. The eventual current is zero: the capacitor charges and blocks steady current in the series RLC circuit.

First construct the periodic response required in the introductory clause. At ${\omega }_0^2=1/(LC)$, the terms $L{I}_0^{\prime \prime }+I_0/C$ cancel for a sinusoid, leaving $R{I}_0^{\prime }={\omega }_0\cos ({\omega }_{0}t)$. ThusThis resonant series RLC response dissipates energy through Joule heating. Over one period,The Q factor, using the maximum magnetic energy $L{I}_M^2/2$, is consequently

Every solution is the resonant series RLC response $I_0(t)=R^{-1}\sin ({\omega }_{0}t)$ plus a solution of the homogeneous damped harmonic oscillator. Define $\gamma =R/(2L)$ and ${\omega }_0^2=1/(LC)$. The three possibilities areThese are respectively the underdamped RLC response, critically damped RLC response and overdamped RLC response. Both exponents in the last case are negative because $0<\kappa <\gamma$. All three homogeneous contributions tend to zero as $t\to \infty$, so every response approaches the unique periodic steady current $I_0(t)$. No initial conditions were specified in this clause; they determine $A,B$.