An alternating current of frequency 60Hz has maximum value of 120A. (1) write down the equation for instateneous value. (2) recording the time from instant the current is zero and becoming positive find the instateneous time after 1/360 second

Physics

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2 parts
a

I=120sin⁡(120πt)I = 120 \sin(120\pi t)

b

603 A60\sqrt{3} \text{ A}

Step-by-step solution

6 steps

Haya Jozee, hebu tutatue hili.

Part a· 3 steps

Andika mlinganyo wa thamani ya papo hapo.

  1. Tambua vigezo vilivyotolewa.

    Tunapewa:
    Thamani ya juu zaidi ya mkondo (maximum current), I0=120 AI_0 = 120 \text{ A}
    Marudio (frequency), f=60 Hzf = 60 \text{ Hz}
    Mkondo unaanzia sifuri na kuwa chanya, hivyo tunatumia kazi ya sine.

  2. Andika mlinganyo wa jumla wa mkondo wa papo hapo.

    Mlinganyo wa mkondo wa papo hapo (instantaneous current) ni:

    I=I0sin⁡(2πft)I = I_0 \sin(2\pi ft)
  3. Badilisha thamani za I0I_0 na ff kwenye mlinganyo.

    I=120sin⁡(2π×60t)I = 120 \sin(2\pi \times 60t) I=120sin⁡(120πt)I = 120 \sin(120\pi t)

    Mlinganyo wa thamani ya papo hapo ni I=120sin⁡(120πt)\boxed{I = 120 \sin(120\pi t)}.

Part b

603 A60\sqrt{3} \text{ A}

Tafuta mkondo wa papo hapo baada ya sekunde 1/360.

  1. Tumia mlinganyo kutoka sehemu (a).

    I=120sin⁡(120πt)I = 120 \sin(120\pi t)
  2. Badilisha thamani ya muda t=1360 st = \frac{1}{360} \text{ s} kwenye mlinganyo.

    I=120sin⁡(120π×1360)I = 120 \sin\left(120\pi \times \frac{1}{360}\right)
  3. Rahisisha na kokotoa thamani ya II.

    I=120sin⁡(120π360)I = 120 \sin\left(\frac{120\pi}{360}\right) I=120sin⁡(π3)I = 120 \sin\left(\frac{\pi}{3}\right)

    Tunajua kuwa sin⁡(π3)=sin⁡(60∘)=32\sin\left(\frac{\pi}{3}\right) = \sin(60^\circ) = \frac{\sqrt{3}}{2}.

    I=120×32I = 120 \times \frac{\sqrt{3}}{2} I=603 AI = 60\sqrt{3} \text{ A}

    Thamani ya mkondo wa papo hapo ni 603 A\boxed{60\sqrt{3} \text{ A}}.

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