Determining the voltage of a power source

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SUMMARY

The discussion focuses on determining the initial voltage, Vt, of a power source in a biological membrane context, with options provided for its value. The time constant is identified as 120 ms, leading to an internal resistance of 12 kΩ. The conversation clarifies that in a series RC circuit, when the switch is closed, the voltage across the resistor starts at Vt and decreases exponentially, while the voltage across the capacitor rises towards Vt. The conclusion drawn is that if the total battery voltage is 40 mV, then the voltage across the capacitor will equal 40 mV when fully charged, resulting in 0 mV across the resistor.

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http://img228.imageshack.us/img228/5797/screenshot20100527at215.png

This is for a biological membrane, but the principles are the same. I am asked what the value of the initial voltage, Vt is.The options are:

What was the value of the initial voltage, Vt?
A. +58 mV
B. -58 mV
C. +40 mV
D. +116 mV
E. It cannot be determined.

I can see that the time constant is 120ms, and hence the internal resistance is 12kohms. However, I am unable to see how I will calculate Vt, as the 40mV initially seen is surely not the potential difference across the battery.

Many Thanks
 
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If this is equivalent to a series RC circuit with a fixed resistor and a switch then when the switch is closed the voltage across the resistor starts at Vt and falls exponentially towards zero whereas the voltage across the capacitor starts at zero and rises towards Vt.
 
Thanks for the reply. If 40mV was the total battery voltage, and there are 40mV across the capacitor, does that mean that there are 0v across the resistor? My physics is really rusty ever since I became a medic..!
 
If 40mV is the battery voltage then the full 40mV will be across the capacitor when it is fully charged and 0mV will be across the resistor.
 
Last edited:
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