The question is:

Assume 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.

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)

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.

 

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

 

That makes sense, thank you for the response

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