Why is there no voltage drop through this resistor?

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The discussion centers on the lack of voltage drop across a resistor in a circuit involving a 9V battery, a capacitor, and an LED. The voltage readings show 6.5V across the capacitor and 1.5V across the LED, but 0V across the resistor. The explanation points to the behavior of capacitors, indicating that the capacitor is likely holding charge and not allowing current to flow through the resistor. Participants emphasize the importance of understanding capacitor function to grasp the circuit's behavior. The thread concludes with a reminder for politeness in future discussions.
trickybilly
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I have wired in series: + of 9V battery --- + of 1000 uF capacitor --- 660 Ohm resistor --- + of a green LED --- - of the 9V battery. Remark: the 9V battery is a bit depleted 7.8 V unconnected.

Voltage drop across the capactor is 6.5V, across the LED is 1.5V but 0V across the resistor. Why there is no voltage drop across the resistor?
 
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Try this:
1) discharge the capacitor by placing the resistor across it for several seconds.
2) connect everything back up except the LED to - of the battery
3) connect the - of the battery while carefully watching the LED.

What did you see?

4) now remove the capacitor

What do you see now?

BoB
 
trickybilly said:
I have wired in series: + of 9V battery --- + of 1000 uF capacitor --- 660 Ohm resistor --- + of a green LED --- - of the 9V battery. Remark: the 9V battery is a bit depleted 7.8 V unconnected.

Voltage drop across the capactor is 6.5V, across the LED is 1.5V but 0V across the resistor. Why there is no voltage drop across the resistor?
Do you understand how capacitors work?

EDIT: Let me rephrase that: you clearly don't yet understand how capacitors work, so that's where you should focus. Study capacitors for a bit and you will understand what is happening in your circuit. That's where rbelli1 is leading you.
 
@phinds I have received my answer. Do YOU understand it?

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trickybilly said:
Do YOU understand it?
You'll bite a helping hand ? Thanks for the warning.
 
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Since the OP has been answered we will close this thread. Everyone, please remember to be polite in future threads.
 
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I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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