The voltage v(t) shown in Fig. P7.85a is given by the graph shown in Fig. P7.85b. If iL(0)...

(b)

Determine power is supplied by the source at \(t=4 s\).

Power supplied by the source is,

$$ \begin{aligned} P_{S}(t) &=v(t) i(t) \\ &=v(t)\left[i_{L}(t)+\frac{v(t)}{R}\right] \end{aligned} $$

But

$$ \begin{aligned} i_{L}(t) &=\frac{1}{L} \int v_{L} d t i_{L}(4) \\ &=\frac{1}{2} \int_{0}^{2} 10 d t+\frac{1}{2} \int_{2}^{4}-10 d t=\frac{1}{2}[10 t]_{0}^{2}-\frac{1}{2}[10 t]_{2}^{4} \\ &=\left(\frac{1}{2}[10(2)]-\frac{1}{2}[10(0)]\right)-\left(\frac{1}{2}[10(4)]-\frac{1}{2}[10(2)]\right) \end{aligned} $$

Simplify further.

$$ \begin{aligned} i_{L}(t) &=10-0-10-0 \\ &=0 \mathrm{~A} \end{aligned} $$

Then

$$ \begin{aligned} P_{S}(t) &=v(t)\left[i_{L}(t)+\frac{v(t)}{R}\right] \\ &=v(t)\left[0+\frac{v(t)}{R}\right] \\ &=\frac{v(t)^{2}}{R} \end{aligned} $$

Simplify further.

$$ \begin{aligned} P_{S}(t) &=\frac{10^{2}}{2} \\ &=\frac{100}{2} \\ &=50 \mathrm{~W} \end{aligned} $$

(c)

Determine current \(i(t)\) at \(t=6 s\).

$$ i(t)=i_{R}+i_{L} $$

At \(t=6 s\) the current through resister is zero.

$$ \begin{aligned} i_{L}(t) &=\frac{1}{L} \int v_{L} d t \\ i_{L}(6) &=\frac{1}{2} \int v_{L} d t \\ &=\frac{1}{2} \int_{0}^{2} 10 d t+\frac{1}{2} \int_{2}^{5}-10 d t \\ &=\frac{1}{2}[10 t]_{0}^{2}-\frac{1}{2}[10 t]_{2}^{5} \end{aligned} $$

Simplify further.

$$ \begin{aligned} i_{L}(6) &=\left(\frac{1}{2}[10(2)]-\frac{1}{2}[10(0)]\right)-\left(\frac{1}{2}[10(5)]-\frac{1}{2}[10(2)]\right) \\ &=10-0-25+10 \\ &=20-25 \\ &=-5 \mathrm{~A} \\ i_{L}(6) &=-5 \mathrm{~A} \end{aligned} $$

Determine total current.

$$ \begin{aligned} i(t) &=i_{R}+i_{L} \\ &=0+(-5) \\ &=-5 \mathrm{~A} \end{aligned} $$

(d)

Determine power is absorbed by inductor.

$$ \begin{aligned} &P_{L}=v(t) i_{L}(t) \\ &i_{L}(3)=\frac{1}{L} \int v_{L} d t \\ &=\frac{1}{2} \int_{0}^{2} 10 d t+\frac{1}{2} \int_{2}^{3}-10 d t \\ &=\frac{1}{2}[10 t]_{0}^{2}-\frac{1}{2}[10 t]_{2}^{3} \end{aligned} $$

Simplify further.

$$ \begin{aligned} &=\left(\frac{1}{2}[10(2)]-\frac{1}{2}[10(0)]\right)-\left(\frac{1}{2}[10(3)]-\frac{1}{2}[10(2)]\right) \\ &=10-0-15+10 \\ &=20-15 \\ &=5 \mathrm{~A} \end{aligned} $$

And

$$ v(3)=-10 $$

Then power absorbed by the inductor is,

$$ \begin{aligned} P_{L}(3) &=v(t) i_{L}(t) \\ &=(-10)(5) \\ &=-50 \mathrm{~W} \\ P_{L}(3) &=-50 \mathrm{~W} \end{aligned} $$