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A battery of E.M.F. & internal resistance 2 ohm is connected to a resistor of 100 ohm resistance through an ammemeter. The resistance of the ammemeter is 4/3 ohm; a voltmeter has also been connected to find the PD across the resistor.
(i) If the ammemeter reads 0.02amp,what is resistance of the voltmeter.
(ii)If the voltmeter reads 1.10V,what is the error in the reading.


[Please Show your calculations & if possible the circuit drawing.]

2006-09-03 19:10:21 · 1 answers · asked by baban 2 in Science & Mathematics Physics

EMF of the battery is 1.4V.
Sorry I forgot to mention it.

2006-09-04 05:26:47 · update #1

1 answers

Didn't you leave out the voltage of the battery?

2006-09-04 03:54:43 · answer #1 · answered by sojsail 7 · 0 0

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