A battery with E = 12.0 V and internal resistance r=1.3 is connected to two 8.5-

Question

A battery with E = 12.0 V and internal resistance r=1.3 is connected to two 8.5-k resistors in series. An ammeter of internal resistance 0.50 measures the current, and at the same time a voltmeter with internal resistance 20 k measures the voltage across one of the 8.5-k resistors in the circuit

Part A What does the ammeter read? Express your answer to two significant figures and include the appropriate units

Part B What does the voltmeter read? Express your answer to two significant figures and include the appropriate units.

Part C

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

Express your answer using two significant figures.

Part D

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

Express your answer using two significant figures.

Explanation / Answer

A)

combined resistance 1/r = 1/(20*10^3) + 1/(8.5*10^3) =1.676*10^-4

r = 5964.91

total resistance R = 8500 +1.3 +5964.91 + 0.5 = 14466.712

Ammeter reads current in the circuit I = V/R = 12 / 14466.712

I = 8.3 *10^-4 A

B)

the voltmeter reads potential difference across the combination of voltmeter and 8.5 k is

potential difference = r*I =  5964.91 *  8.3 *10^-4 = 4.9 V

C)

without meters

total resistance R = 1.3 + 8500 =8501.3

current I = V / R = 12 / 8501.3

I = 1.411 *10^-3

% "error" =[( 1.411 *10^-3 - 8.3 *10^-4  ) /1.411 *10^-3 ]*100 = 41%

D)

% "error" from the voltage =[ (12- 4.9)/12]*100 = 59 %

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