Voltage across components in a zero-current circuit

[Yh Hoo]

Hi i am having some obstacles in building up my concept in dc circuit


First of all, as we know, voltmeter is made up of a series circuit with both ends connected to two points of other circuits, while inside the series circuit, a galvanometer and resistor is connected in series. How the voltmeter measure the potential difference?? is it that the reading appears on the voltmeter directly equal to the potential difference across the built-in resistor with very very high resistance inside the voltmeter?


Now for the circuit B, why will the reading of voltmeter be equal to emf of the cell?? supposedly there should be no potential difference across the built-in resistor in the voltmeter right?

But for the circuit A, what should be the potential difference across the diode?? in this case there should be potential difference across the diode right? since the emf is having tendency to do work against the built-up opposing voltage in the diode.
Now, what should the voltmeter read ? and the potential difference across the internal resistor should be what?

Hope you can guide me.

 

[Runei]

In both cases you would read the voltage across the battery (if we talk ideally).
There will be a small leakage current through the reverse biased diode, but for any practical idea - voltage in both cases is the emf of the battery.

 

[Windadct]

Do not let the internal resistance of the meter confuse you. For this type of meter - it is really a Current meter - and the resister establishes a ratio of current vs the voltage applied. In reality - yes the voltage is across this resistor - BUT this is part of a system that makes a voltage meter -the meter is not truly measuring the voltage across the internal resistor.
It is much better to break these cases into simpler parts - consider Just the meter and understand that - then when you use it to measure diodes and the like - do not get distracted by what is going on "inside" the meter.
The dynamics of the battery and the Diode are actually more complicated than the meter - if you try to figure it all out at once - you will be perplexed indeed.

 

[Yh Hoo]

Do not let the internal resistance of the meter confuse you. For this type of meter - it is really a Current meter - and the resister establishes a ratio of current vs the voltage applied. In reality - yes the voltage is across this resistor - BUT this is part of a system that makes a voltage meter -the meter is not truly measuring the voltage across the internal resistor.
It is much better to break these cases into simpler parts - consider Just the meter and understand that - then when you use it to measure diodes and the like - do not get distracted by what is going on "inside" the meter.
The dynamics of the battery and the Diode are actually more complicated than the meter - if you try to figure it all out at once - you will be perplexed indeed.

Thanks for your guide. Another question,for caseA,what is the potential difference across the internal resistance n the
Diode??

 

[alphysicist]

Hello,

I can definitely understand your confusion and obstacles in understanding DC circuits. Let's break down the concepts one by one to help build up your understanding.

Firstly, regarding the voltmeter, you are correct that it is made up of a series circuit with a galvanometer and a high resistance resistor. The galvanometer measures the current passing through the circuit, and the high resistance resistor limits the current flow so that it does not interfere with the circuit being measured. The reading on the voltmeter is indeed equal to the potential difference across the built-in resistor, as this is what the galvanometer is measuring.

In circuit B, the voltmeter reads the emf of the cell because there is no current flowing through the circuit. In a zero-current circuit, the voltage across each component is equal to the emf of the cell. This is because there is no voltage drop across any of the components, so the voltage measured by the voltmeter is equal to the voltage supplied by the cell.

In circuit A, the potential difference across the diode depends on the direction of current flow. If the current is flowing in the forward direction, there will be a small voltage drop across the diode due to its internal resistance. However, if the current is flowing in the reverse direction, the diode will act as an open circuit and there will be no potential difference across it. The voltmeter will read the same potential difference as in circuit B, which is equal to the emf of the cell.

I hope this helps clarify some of your doubts. It is important to remember that in a zero-current circuit, the voltage across each component is equal to the emf of the cell. Keep practicing and experimenting with circuits to build your understanding. Good luck!