Electric Current — Question 6

Section 26-1 · Physics pages 4162–4163

Question 6 A shaped pulse fixes its peak from a specified transported charge A unidirectional pulse obeys I(t) = Ip sin(πt/T) for 0 ≤t ≤T and is zero otherwise. It must deliver positive charge Q0. t I Axes and labels specify the model; the sketch is schematic. (a) Find the required peak current. (b) Find the fraction delivered by t = T/3 and the time when half is delivered.
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Question 6 - Solution A shaped pulse fixes its peak from a specified transported charge t I (a) Method and calculation Q0 = ∫T 0 Ip sin(πt/T) dt = 2IpT/π. Ip = πQ0/(2T). (b) Interpretation and further calculation Q(t)/Q0 = 1 2[1 −cos(πt/T)], Q(T/3)/Q0 = 1/4. Q(t1/2)/Q0 = 1/2 ⇒t1/2 = T/2. Physical check The mean pulse current is smaller than its peak by the factor 2 divided by pi.

Question 6

A shaped pulse fixes its peak from a specified transported charge

A unidirectional pulse obeys I(t)=Ipsin⁡(πt/T)I(t)=I_p\sin(\pi t/T) for 0≤t≤T0\le t\le T and is zero otherwise. It must deliver positive charge Q0Q_0.

Diagram shown in the original SVG worksheet.

  1. Find the required peak current.

  2. Find the fraction delivered by t=T/3t=T/3 and the time when half is delivered.

Question 6 - Solution

A shaped pulse fixes its peak from a specified transported charge

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Q0=∫0TIpsin⁡(πt/T)dt=2IpT/π.Q_0=\int_0^T I_p\sin(\pi t/T)\,dt=2I_pT/\pi.Ip=πQ0/(2T).I_p=\pi Q_0/(2T).

(b) Interpretation and further calculation

Q(t)/Q0=12[1−cos⁡(πt/T)],Q(T/3)/Q0=1/4.Q(t)/Q_0=\tfrac12[1-\cos(\pi t/T)],\quad Q(T/3)/Q_0=1/4.Q(t1/2)/Q0=1/2⇒t1/2=T/2.Q(t_{1/2})/Q_0=1/2\Rightarrow t_{1/2}=T/2.

Physical check

The mean pulse current is smaller than its peak by the factor 2 divided by pi.