A battery with E = 12.0 V and internal resistance r=1.9? is connected to two 7.5

Question

A battery with E = 12.0 V and internal resistance r=1.9? is connected to two 7.5-k? resistors in series. An ammeter of internal resistance 0.60 ? measures the current, and at the same time a voltmeter with internal resistance 20 k? measures the voltage across one of the 7.5-k? resistors in the circuit.

Part A

What does the ammeter read? correct answer = Isource = 9.26×10?4 A

Part B

What does the voltmeter read? correct answer = Vmeter = 5.05 V

Part C

What is the % "error" from the current without meters?

Explanation / Answer

PART A) Ammeter is connected in series with resistor and voltmeter connected across one of the 7.5 k-ohm resistor

so total resistance in the circuit R= 1.9 + 0.60 +7500 + (7.5*20/27.5) = 12952.5 ohm

current in the circuit i 1= E/R = 12/12952.5 = 9.26*10^-4 A

PART B) current through the 7.5 k-ohm resistor branch across which voltmeter is connected

is = 20*9.26*10^-4/27.5 = 6.7345*10^-4 A

so reading of voltmeter = 6.7345*10^-4*7500 = 5.05 v

PART C) current without meters i2= 12/(1.9 + 0.60 + 15000) = 7.99*10^-4 A

error = (i1 - i2) /i2 = (9.26 - 7.99)*10^-4/(7.99*10^-4 ) = 0.158

%error = 15.8 %

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