Question

Tolong dibantu no 4 tolong pakai cara yang lengkap dan jelas . Tolong jawabnya yang benar

Answer 1

Verified answer

No. 4
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Diket :
Arus, I = (2,20 +/- 0,1) ampere
Maka :
I = 2,20 ampere
ΔI = 0,1 ampere

Hambatan , R = (10,0 +/- 0.2) ohm
Maka :
R = 10,0 ohm
ΔR = 0,2 ohm

Tanya :
Penulisan laporan daya, P = __?

Jawab :
Step 1
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P = I².R
P = 2,20²•10,0
P = 48,4 watt

Step 2
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P = I².R
dP/dI = 2.I.R
dP/dI = 2•2,20•10,0
dP/dI = 44,0

Step 3
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P = I².R
dP/dR = I²
dP/dR = (2,20)²
dP/dR = 4,84

Step 4
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ΔP = √[(dP/dI.ΔI)² + (dP/dR.ΔR)²]

ΔP = √[44,0•0,1]² + [4,84•0,2]²
ΔP = √[4,4² + 0,968²]
ΔP = √(19,36 + 0,937024)
ΔP = √(20,297024)
ΔP = 4,5 watt

Jadi penulisan daya nya adalah

P = (P +/- ΔP)
P = (48,4 +/- 4,5) watt

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