# Question from text book

Discussion in 'Homework Help' started by gte, Feb 5, 2010.

1. ### gte Thread Starter Active Member

Sep 18, 2009
347
4
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?

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
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.

3. ### gte Thread Starter Active Member

Sep 18, 2009
347
4
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)

4. ### n1ist Active Member

Mar 8, 2009
171
16
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

5. ### gte Thread Starter Active Member

Sep 18, 2009
347
4
That makes sense, thank you for the response