Question 1:
The output (Vout) of the voltage divider shown in the left figure below is to be measured with voltmeters having input resistances of 1 MΩ, 10 kΩ, and 667 Ω. What voltage will each meter indicate? (Only one meter is connected at a time.) Most digital voltmeters nowadays have input resistances around 10 MΩ: maybe now you see why such a large value is helpful. Would you want a current meter to have a very large or a very small input resistance?

To find the voltage drop we want to consider what the resistance of the 2K resistor is in parallel with the resistor from the voltmeter
i)                \(\frac{1}{r_{tot}}\) = \(\frac{1}{2k}\) +\(\frac{1}{1M}\) ;    \(\frac{1}{r_{tot}}\) = 1996 \(\Omega\)
ii)               \(\frac{1}{r_{tot}}\) =  \(\frac{1}{2k}\) +\(\frac{1}{10k}\) ; \(\frac{1}{r_{tot}}\)  =  1666  \(\Omega\)
iii)               \(\frac{1}{r_{tot}}\) =  \(\frac{1}{2k}\) + \(\frac{1}{667}\)\(\frac{1}{r_{tot}}\)  = 500  \(\Omega\)
Now we need to figure out what the voltage drop is across two resistors in parallel.
 
\(V_{drop}=V_0\cdot\frac{R_{tot}}{\left(R_{tot}+R_1\right)}\)
Note: if \(R_{voltmeter}\) = infinity our expected voltage drop would be 4 volts.
i)                 6V*1996/(1996+1000) = 3.997 V
ii)                6V*1666/(1666+1000) = 3.749 V
iii)               6V(500/1500) = 2V