An ammeter whose internal resistance is 65 reads 5.25 mA when connected in a circuit containing a battery and two resistors in series whose values are 640 and 440 . What is the actual current when the ammeter is absent?
answer should be in mA
2007-02-14 15:13:43 · 7 answers · asked by ZG786 1 in Science & Mathematics ➔ Physics
LOL. Wait till they put them in parallel.
E = I x R = 5.25 x (640+440+65) = 6.011v
I = E/R = 6.011/1080
I = 5.56mA
2007-02-14 15:43:11 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The resistance of ammeter = 65 Ohm
Total resistance in the circuit = 640 + 440 + 65 = 1145 Ohm
Since current = 5.25mA we can find the applied voltage
V = I X R = 5.25 mA X 1145 = 6.01125 V
So if ammeter is absent the resistance = 640+440 = 1080 Ohm
1080 connected to 6.01125 V will give 6.01125/1080 = 5.565972 mA
2007-02-14 15:22:15 · answer #2 · answered by Thomas Jude 2 · 0⤊ 0⤋
Insufficient information is given. The internal resistance of the battery is also needed. Usually batteries have very low Resistance so I will give an answer assuming zero resistance in the battery. The total resistance in the original circuit is 640+440+65=1145 ohms
v=I*R
the voltage of the battery is 1145*5.25/1000=6.01125volts
The total resistance without the amp meter is 640+440=1080
I=V/R
6.01125/1080=5.565972222 mA
Any instructor that allows you to specify this many digits in an answer should not be trusted. Since the input data was mostly 2 significant figures the answer should be rounded to at most 3 significant digits.
2007-02-14 15:32:29 · answer #3 · answered by anonimous 6 · 0⤊ 0⤋
Do you know Ohm's law? I = V/R
This should be easy since everything is in series.
Resistors in series add. Simple enough. Without the amp meter, the total resistance that the battery "sees" is the two resistors only. You should be able to take it from there.
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Hint:
With the amp meter: V(battery) = I * (65 + 640 +440)
2007-02-14 15:20:38 · answer #4 · answered by Randy G 7 · 0⤊ 1⤋
What Is Absent
2016-09-30 10:15:19 · answer #5 · answered by Anonymous · 0⤊ 0⤋
0.0055659722222222222222222 amps
or 5.5659 ma
(Batt-volt) / (640+440+65) = .00525 amps
batt-volt = 6.01125
with no ammeter: batt-volt(above)/ (640+440) = long figure at top, etc.
(All from ohms law: amp = volts/resistance)
2007-02-14 15:26:37 · answer #6 · answered by answerING 6 · 0⤊ 0⤋
About 5.565972 mA
2007-02-14 15:18:55 · answer #7 · answered by Atomic Collision 2 · 0⤊ 0⤋