Question 5.11: Figure 5.23 shows a voltage waveform of a 0.5 H inductor. De......

v(t)={\frac{10}{2}}t=5t \ {V}     0\lt t\lt 2        (5.164)

\nu(t)=10\,\mathrm{V}\quad2\lt t\lt 4        (5.165)

i=\frac{1}{L}\int\nu\,\mathrm{d}t=\frac{1}{0.5}\left[\int\limits_{0}^{2}{5tdt}+\int\limits_{2}^{4}{10dt} \right]        (5.166)

i=\frac{1}{0.5}\left[\frac{5\times(4-0)}{2}+10(4-2)\right]=60\,\mathrm{A}       (5.167)