one MC about circuit concept?
Why is the answer B?
there should be nearly no current passes through the light bulb so the potential drop is 0V. There should be 0 resistance in the ammeter and the potential drop is also 0 V. So consider the loop joining the light bulb, the voltmeter and ammeter, shouldn't the voltmeter detects 0V?
I am not very familiar with voltmeter, can you tell me when the voltmeter detect 0 voltage and what is the principle behind it. (Briefly)
Thanks for helping me
- 天同Lv 74 年前最愛解答
What you are saying is correct only if both the ammeter and voltmeter are ideal. But the question doesn't tell that the two are ideal meters. Since the two meters are used in an experiment, we cannot take it for granted that they are ideal. Hence, the internal resistances of the meters need to be considered in answering the question.
As a practical voltmeter has a large resistance, the current flows through the bulb and voltmeter branch (you may just take the voltmeter as a resistor of very high resistance) is extremely small as compared with the current that passes through the ammeter branch, because of the low resistance of the ammeter.
Therefore, statement (1) is right because the current passing through the bulb-voltmeter branch is small.
Statement (2) is right, because a large current flows through the ammeter branch (due to the small resistance of the ammeter).
Statement (3) is wrong. The bulb-voltmeter branch is connected to the battery. Thus the potential difference across this branch equals to the battery emf (assume internal resistance of battery is negligible). Because the bulb has resistance much much lower than that of the voltmeter, most of the potential drop will be across the voltmeter (as said above, just imagine the voltmeter as a resistor of high resistance). A voltmeter measures the potential drop across its two terminals (or the potential drop across its own internal resistance), the voltage thus recorded by the voltmeter will be nearer equal to the emf of the battery.
Points to note:
Do not mistakenly think that a low current must give rise to a low potential drop, and a large current leads to a high potential drop. You need to consider the involved resistances in the circuit too.
In this problem, although the current in the bulb-voltmeter branch is small, the potential difference across this branch is not near zero, because of the high resistance of the voltmeter. Similarly, the current in the ammeter branch is large, but the potential difference across this branch equals to that of the other branch (but NOT higher than). Both branches have potential differences equal to the battery emf.