a circuit consists of a series combination of 6.00 ???k?? and 4.50 ???k?? resistors connected across a 50.0-v battery having negligible internal resistance. you want to measure the true potential difference (that is the potential difference without the meter present) across the 4.50 ???k?? resistor using a voltmeter having an internal resistance of 10.0 k??.
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21.428volts when the meter is not present.
at first 10k is parallel with 4.50k
that is 4.50*10/(4.5+10)=3.103ohms
the circuit is 3.103ohms series with 6k across 50v. therefore 50*3.103/(3.103+6)=17.043v
true potential difference is 50*4.5/(4.5+6)=21.428v
at first 10k is parallel with 4.50k
that is 4.50*10/(4.5+10)=3.103ohms
the circuit is 3.103ohms series with 6k across 50v. therefore 50*3.103/(3.103+6)=17.043v
true potential difference is 50*4.5/(4.5+6)=21.428v
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