Past exam of the physics 2 paper of the University of Oxford / 2016 resit / 11

Initially the open switch leaves a series capacitor-resistor-inductor circuit driven by the battery. After a long time the electric current is zero and the capacitor voltage is $V_0$. Closing the switch bypasses and, under the stated idealization, instantaneously discharges the capacitor. The inductor current remains continuous at zero.

The remaining RL circuit obeys $L\dot{I}+RI=V_0$, soThe capacitor's impulsive discharge through the ideal short is distinct from the finite electric current through the coil. A finite switch resistance would resolve that separate fast transient.

Put $t_1=0.005\mathrm{s}$, $\tau =t-t_1$, and $I_1=2(1-e^{-1.875})$. When the switch reopens, the circuit again contains the battery and all three series elements. Inductor current continuity and capacitor voltage continuity giveThus $I^{\prime }(0^{+})=(V_0-R{I}_1)/L\simeq 115.0\mathrm{A}{\mathrm{s}}^{-1}$, not zero. Differentiating the loop equation gives the homogeneous current equationThe battery is still connected; its constant voltage disappears on differentiating, and the eventual capacitor voltage is $V_0$.

With $\alpha =R/(2L)=187.5{\mathrm{s}}^{-1}$ and ${\omega }_0=1/\sqrt{LC}$, the inequality $\alpha <{\omega }_0$ makes the circuit underdamped. The damped angular frequency and ordinary frequency areThe full current satisfying both initial conditions isIts amplitude envelope decays as $e^{-\alpha \tau }$. ThereforeThis is measured after reopening; on the clock starting with the first closure it is $t\simeq 0.02098\mathrm{s}$. An instantaneous current zero is not an amplitude decay measurement.