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- Thread starter harshakantha
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- #27

Femme_physics

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I took on solving it myself and got 5.22V (I realized drawing "I2" was a mistake)

http://img32.imageshack.us/img32/3595/voltagedrop.jpg [Broken]

http://img32.imageshack.us/img32/3595/voltagedrop.jpg [Broken]

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- #28

ehild

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You are welcome.. And one more little hint: When doing calculations, do not round the data too early, too much. You rounded the current to two significant digits so the error of the result is 0.1 V. During calculations, keep at least one more significant digits as needed in the final result.

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- #29

ehild

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I took on solving it myself and got 5.22V (I realized drawing "I2" was a mistake)

Good job! Congratulations!

Just a little hint, the same I said to harshakantha: Do not round off too much during the calculations. That current was 0.00647 with 3 digits. Both of you rounded it to 0.0065 which caused an error of 0.066 V in the voltage in harshakantha's calculation as he multiplied it with the bigger resistance. You used the smaller resistance, so your error is one magnitude less, 0.004 V.

ehild

- #30

Femme_physics

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Thank you ehild, so encouraging! :)

You're right I had a feeling I was rounding too much.

You're right I had a feeling I was rounding too much.

- #31

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Thanks ehild for your advice I'll keep it on my mind, bye...

- #32

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Oh.. thanks Femme_physics for your post, hope to get help again from both of you in future. bye..

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