Question

A $1.00mH$ inductor and one $1.00\mu F$ capacitor are connected in series.The current in the circuit is described by $I=20t$ where t is in seconds and I is in amperes. Initially, the capacitor has no charge. Determine
a. The voltage across the inductor as a function of time,
b.The voltage across the capacitor as a function of time
c.The time when the energy stored in the capacitor first exceeds that in the inductor.

Answer 1

a) $I=20t$

$V_{inductor}=L\frac{dI}{dt}=1\times {10}^{-3}\times 20=2\times {10}^{-2}V$

b) $Q={\int }_0^{t}Idt\Rightarrow Q=10t^2$

$V_{capacitor}=\frac{Q}{C}=\frac{10t^2}{{10}^{-6}}={10}^7{t}^{2}V$

c) Energy in capacitor $=\frac{1}{2}\times C\times V^2$

$U_C=\frac{1}{2}\times {10}^{-6}\times {10}^{14}{t}^4=\frac{1}{2}\times {10}^8{t}^4$

$U_{inductor}=\frac{1}{2}\times L\times I^2=\frac{1}{2}\times 1\times {10}^{-3}\times 400t^2$

$=\frac{1}{2}\times \frac{4\times t^2}{10}$

$U_{cap}=U_{ind}$

$\Rightarrow =\frac{1}{2}\times {10}^8\times t^4=\frac{1}{2}\times \frac{4}{10}\times t^2$

$t=2\sqrt{10}\times {10}^{-5}s$