The question is: So 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?
E=battery voltage r1=1 Mohm r2=meter input impedance E=Vr1+Vr2=E/2+Vr2 Thus, Vr2=E-E/2=E/2=Vr1 Vr1=I*r1 Vr2=I*r2 since I is the same through both resistances Vr1/r1=Vr2/r2 since Vr1=Vr2 r1=r2=1 Mohm An alternative way to solve it is by using the voltage divider rule.
Hi Mik3, Thanks for the response, I'm still a little confused with the equations having multiple = signs in them, could your equation also be stated as the following? How can I plug real numbers into it with multiple equals in the equation? 2E = I * R2 E = I * (R1+R2)
The multiple equals just means that all things are equal. E=Vr1+Vr2=E/2+Vr2 is the same as E=Vr1+Vr2 and E=E/2+Vr2 Just view this problem as a simple voltage divider. You have a power supply and two resistors in series. One is the 1M resistor. The other is the input impedance of the meter. You know the voltage across the meter, so solve for its resistance. /mike