A battery with E = 12.0 V and internal resistance r=1.2 is connected to two 6.5-

Question

A battery with E = 12.0 V and internal resistance r=1.2 is connected to two 6.5-k resistors in series. An ammeter of internal resistance 0.60 measures the current, and at the same time a voltmeter with internal resistance 18 k measures the voltage across one of the 6.5-k resistors in the circuit. What does the ammeter read? What does the voltmeter read? What is the % "error" from the current without meters? What is the % "error" from the voltage without meters? I figured the other ones out except for part D, the % error from the voltages. Help me please.

Explanation / Answer

We have to calculate net resistance of the circuit , without voltmeter
All the resistane will be in the series = 1.2+0.6+2*6.5*103 = 13.001 k-ohm
Current in the circuit = V/R = 12/13.001 = 9.23*10-4 ampere
Voltage in the resistance 6.5 k-ohm = 9.23*10-4*6.5*103 = 5.9995 volt = 6 V
Now we consider the resistance of the voltmeter,
6.5 kohm and rv = 18 kOhm are in parallel
Therefore Rnet = 4.78 kOhm
now all the resistance with Rnet will be in series
Requivalent   = 4.78 + 6.5 + (0.6+1.2)*10-3 = 11.28 kOhm
So current in the circuit
I = V/R = 1.06 4*10-3 ampere
Potential in the 6.5kohm and voltmeter will be same
let "i" is the current in the 6.5 kohm , therefore
i*6.5 = (1.06 4*10-3 - i)18
i = 7.78*10-4 ampere
Voltage in the 6.5 kohm = 5.06 volt
Percentage error =( 6-5.06 / 6)*100 = 15.67 %

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