Past exam of the physics 2 paper of the University of Oxford / 2014 / 9

The source diagram has $2R$ in series with two parallel branches: one containing $R$ then $L$, and the other containing $2R$. Let their common voltage be $u(t)$, measured from the upper node to the lower return. Kirchhoff's current law and the branch laws giveUsing $I_2=u/(2R)$ in the source equation yields $u=V/2-R{I}_1$. Substitution into the coil branch givesThe switch was open for a long time, so $I_1(0^{-})=0$. Inductor current continuity gives $I_1(0^{+})=0$ for a finite applied voltage.

The constant particular solution is $A=V/(4R)$, and the homogeneous equation has decay rate $2R/L$. Imposing $I_1(0)=0$ givesAs a separate check, the Thévenin equivalent seen by $L$ has open-circuit voltage $V/2$ and resistance $R+(2R\parallel 2R)=2R$, again giving $\tau =L/(2R)$.

Put $e_t=e^{-t/\tau }$. From $u=V/2-R{I}_1$ and $I_0=I_1+I_2$,At $t=0^{+}$ the three values are $I_1=0$ and $I_2=I_0=V/(4R)$. At late times they are $I_1=V/(4R)$, $I_2=V/(8R)$, and $I_0=3V/(8R)$. Thus $I_1$ and $I_0$ increase monotonically while $I_2$ decreases; every transient has the same time constant. By $10\tau$, the exponential correction is only $e^{-10}\simeq 4.54\times 10^{-5}$ of its initial amplitude.

Only the ideal inductor stores energy in this circuit; ideal resistors dissipate energy as heat. With $I_1(\infty )=V/(4R)$,The question excludes the battery's internal energy, and the continuously generated resistor heat is not stored electromagnetic energy.