A battery with epsilon = 12.0 V and internal resistance r = 1.0 Ohm is connected

Question

A battery with epsilon = 12.0 V and internal resistance r = 1.0 Ohm is connected to two 5.0-k Ohm resistors in series. An ammeter of internal resistance 0.40 Ohm measures the current, and at the same time a voltmeter with internal resistance 15 k Ohm measures the voltage across one of the 5 O-k Ohm resistors in the circuit A What does the ammeter read? Express your answer to two significant figures and include the appropriate units. What does the voltmeter read? Express your answer to two significant figures and include the appropriate units. What is the % "error" from the current without meters? Express your answer using two significant figures. What is the % "error" from the voltage without meters? Express your answer using two significant figures.

Explanation / Answer

Part A )

R = combination of resistance of voltmeter and 5 kohm resistor = 15 x 5 /(15 + 5) = 3.75 kohm = 3750 ohm

Ra = resistance of ammeter = 0.40 ohm

Current using ohm's law is given as

i = E/(r + 5000 + 0.40 + 3750) = 12/(1 + 5000 + 0.40 + 3750) = 0.0014 A

Part B)

V = Voltage across Voltmeter = E - i (r + 5000 + 0.40) = 12 - (0.0014) (1 + 5000 + 0.40) = 4.998 volts

Part c)

without meter , current is given as

i = E/(1 + 5000 + 5000) = 12 / (1 + 5000 + 5000) = 0.0012

%age error = (0.0014 - 0.0012) (100) / 0.0012 = 16.7 %

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