It would be a high pitched sound. Hope this helps.
The EMF of the battery includes the force to to drive across its internal resistance. the total resistance:
R = internal resistance r + resistance connected rv
R = r + rv
Now find the current:
V 1= IR
I = R / V1
find the voltage at the battery terminal (which is net of internal resistance) using
V 2= IR
So the voltage at the terminal is:
V = V2 - V1
This is the potential difference vmeter measured by the voltmeter.
Nitrogen isotopes don't have a charge.
Technically, we don't have the information needed to calculate the current,
because you haven't mentioned the units of the 3.5 .
Since the 3.5 is a resistance, we strongly suspect ... and we'll therefore
assume ... that the 3.5 has the units of ohms. Then . . .
Current = (voltage) / (resistance) = (1.5/3.5) = <em>3/7 of an Ampere</em>.
(429 mA, rounded)
