Potential Divider question
Why is C the correct answer for this question?
I know that the voltmeter reading will decrease due to the change (decrease) in resistance of the potential divider.
But why does the brightness of the lamp remain the same?
If the potential divider takes up less amount of voltage, then more voltage should go to the lamp making it brighter?
I know that the voltmeter reading will decrease due to the change (decrease) in resistance of the potential divider.
But why does the brightness of the lamp remain the same?
If the potential divider takes up less amount of voltage, then more voltage should go to the lamp making it brighter?
(edited 11 months ago)
Assuming the voltmeter has infinite resistance, then regardless of with two points u connect the voltmeter to, it will not change the current flowing in the ckt initially as voltmeter will not draw any current.
Original post by S0303
Why is C the correct answer for this question?
I know that the voltmeter reading will decrease due to the change (decrease) in resistance of the potential divider.
But why does the brightness of the lamp remain the same?
If the potential divider takes up less amount of voltage, then more voltage should go to the lamp making it brighter?
I know that the voltmeter reading will decrease due to the change (decrease) in resistance of the potential divider.
But why does the brightness of the lamp remain the same?
If the potential divider takes up less amount of voltage, then more voltage should go to the lamp making it brighter?
The total series resistance between points P and Q does not change.
As the previous poster points out, the voltmeter virtually infinite resistance so no current will flow through it for all practical purposes.
Therefore, the current through the lamp does not change irrespective of the potentiometer wiper position.
The voltmeter will still measure the change in voltage across the potential divider as the wiper changes position.
(edited 11 months ago)
Original post by S0303
Why is C the correct answer for this question?
I know that the voltmeter reading will decrease due to the change (decrease) in resistance of the potential divider.
But why does the brightness of the lamp remain the same?
If the potential divider takes up less amount of voltage, then more voltage should go to the lamp making it brighter?
I know that the voltmeter reading will decrease due to the change (decrease) in resistance of the potential divider.
But why does the brightness of the lamp remain the same?
If the potential divider takes up less amount of voltage, then more voltage should go to the lamp making it brighter?
Are you having an exam soon? If yes, I suspect you cannot figure out is due to exam stress and overlook something simple.
Voltmeter (if ideal) measures the potential difference between 2 points NOT affecting the potential difference.
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