he voltage pulse described by the following equations is impressed across the terminals of a 0.5 capacitor: v{(t) }={0, t≤0s ;4tV 0 s ≤ t≤ 1 s;4e^-(t-1) V, t≥ 1 s} a) Derive the expressions for the capacitor current, power, and energy. b) Sketch the voltage, current, power, and energy as function

Question

he voltage pulse described by the following equations is impressed across the terminals of a 0.5 capacitor:

[latex]v_{\left(t ight) }=\begin{Bmatrix} 0,& t \leq 0 s; \ 4 t V , & 0s\leq t \leq 1s \ 4e^{-\left(t-1 ight) } V, & t\geq 1 s.\end{Bmatrix}[/latex]

a) Derive the expressions for the capacitor current, power, and energy.

b) Sketch the voltage, current, power, and energy as functions of time. Line up the plots vertically.

c) Specify the interval of time when energy is being stored in the capacitor.

d) Specify the interval of time when energy is being delivered by the capacitor.

e) Evaluate the integrals

[latex]\int_{0}^{1 }{p dt } and \int_{1}^{ \infty}{p dt }[/latex]

and comment on their significance

Accepted Answer

a) From Eq. 6.13,[latex]i=C\frac{dv}{dt} [/latex],

[latex]i=\left\{\begin{array}{ll} \left(0.5 imes 10^{-6} ight)(0)=0, & t<0 \mathrm{~s} \ \left(0.5 imes 10^{-6} ight)(4)=2 \mu \mathrm{A}, & 0 \mathrm{~s}1 \mathrm{~s} \end{array} ight.[/latex]

The expression for the power is derived from Eq. 6.16: [latex]p=vi=Cv\frac{dv}{dt} [/latex],

[latex]p=\left\{\begin{array}{ll} 0, & t \leq 0 s \ (4 t)(2)=8 t \mu \mathrm{W}, & 0 \mathrm{~s} \leq t<1 \mathrm{~s} \ \left(4 e^{-(t-1)} ight)\left(-2 e^{-(t-1)} ight)=-8 e^{-2(t-1)} \mu \mathrm{W}, & t>1 \mathrm{~s} \end{array} ight.[/latex]

The energy expression follows directly from Eq. 6.18: [latex]W=\frac{1}{2}Cv^{2} [/latex],

[latex]w=\left\{\begin{array}{ll} 0, & t \leq 0 \mathrm{~s} \ \frac{1}{2}(0.5) 16 t^{2}=4 t^{2} \mu \mathrm{J}, & 0 \mathrm{~s} \leq t \leq 1 \mathrm{~s} \ \frac{1}{2}(0.5) 16 e^{-2(t-1)}=4 e^{-2(t-1)} \mu \mathrm{J}, & t \geq 1 \mathrm{~s} \end{array} ight.[/latex]

b) Figure 6.11 shows the voltage, current, power, and energy as functions of time.

c) Energy is being stored in the capacitor whenever the power is positive. Hence energy is being stored in the interval 0-1 s.

d) Energy is being delivered by the capacitor whenever the power is negative. Thus energy is being delivered for all t greater than 1 s.

e) The integral of p dt is the energy associated with the time interval corresponding to the limits on the integral. Thus the first integral represents the energy stored in the capacitor between 0 and 1 s, whereas the second integral represents the energy returned, or delivered, by the capacitor in the interval 1s to[latex] \infty [/latex] :

[latex]\int_{0}^{1 }{p dt }=\int_{0}^{1 }{8t dt }= 4t^{2}\mid^{1}_{0}= 4\mu J[/latex],

[latex]\int_{1}^{ \infty}{p dt }=\left(-8e^{-2\left(t-1 ight) } ight) dt=\left(-8 ight) \frac{e^{-2\left(t-1 ight) }}{-2}\mid^{\infty}_{1}=-4 \mu J[/latex].

The voltage applied to the capacitor returns to zero as time increases without limit, so the energy returned by this ideal capacitor must equal the energy stored.

Question 6.4: he voltage pulse described by the following equations is imp...

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