The voltage across a lamp is (6.0+-0.1) V and the current flowing through it is (4.0+-0.2) A. Find the power consumed with maximum permissible error in it.
[ Hint: (power =V×I),take the upper and lower limits and find the maximum of the two.]
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Power = VIP = 6*4 = 24 WattdP = P(dV/V + dI/I)dP = 24*(0.1/6 + 0.2/4)dP = 1.6Power consumed ..
Explanation:
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