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