When a current i flows through a conductor of resistance R, the power P consumed is given by P
= i
2R. If the current is measured as (3.50 + 0.05) mA and R = (34.5 + 2%), Find the percentage
uncertainty in P
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P = I^2 × R
= (3.50 + 0.05 mA)^2 × (34.5 + 2%)ohm
= (3.55 mA)^2 × (34.5 +0.69) ohm
= (0.00355 A)^2 × 35.19 ohm
= 0.000443481975 watt(W)
= 0.0004%
I have done it but I don't know whether it is correct or not...
If its correct mark my answer as a brainliest answer plz
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