A battery with epsilon = 12.0 V and internal resistance r = 1.9 ohm is connected

Question

A battery with epsilon = 12.0 V and internal resistance r = 1.9 ohm is connected to two 7.0-kohm resistors in series. An ammeter of internal resistance 0.70 ohm measure the current, and the same time a voltmeter with internal resistance 18 k Ohm measures the voltage across one of the 7.0.kohm resistors in the circuit. 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.

Explanation / Answer

resistance of 7 kohm and 18 kohm will be in parallel so eq resistance = 5040 ohm

now 5040 ohm and 0.7 ohm and 1.9 ohm will be in series so over all resistance

R = ( 5040 + 0.7+1.9) = 5042.6 ohm

now current i = 12/5042.6

i = 2.37 m Amp

so ameter reading = 2.38 mili amp ans part A

so voltmeter reading = ( 12 - 0.7i-1.9i) = 11.9971 V

if ameter is not add then

Req = 5040 +1.9 = 2.38 amp

so % error = 0.42 %

now if voltmeter is now connected the

Req = 7000 + 0.7+1.9 = 7002.6 ohm

now i = 20/7002.6

i = 2.83 m Amp

now votage accros R = ( 12 - 2.6 i) = ( 12- 2.6 * 2.83 *10-3) = 11.992 V

% change = 0.0425 % answer

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