The question is:
2E = I * R
E = I * (R + 1Mohm)
^ Is that correct so far? I was trying to plug in values to see if this actually worked, but I'm not having any luck?
10v = .00001A * 1Mohm
20v = .00001A * 2Mohm
Should the current be a constant in this? Cause when you add resistance in the equation, the voltage number goes up, not down, if the current stays the same? What am I doing wrong?
SoAssume you measure a power supply voltage through a series 1 Mohm resistor and observe that it dropped by 1/2 its former value when no resistor is in series. What is the input impedance of the meter? Explain your answer.
2E = I * R
E = I * (R + 1Mohm)
^ Is that correct so far? I was trying to plug in values to see if this actually worked, but I'm not having any luck?
10v = .00001A * 1Mohm
20v = .00001A * 2Mohm
Should the current be a constant in this? Cause when you add resistance in the equation, the voltage number goes up, not down, if the current stays the same? What am I doing wrong?