Answer:
circuit sketched in first attached image.
Second attached image is for calculating the equivalent output resistance
Explanation:
For calculating the output voltage with regarding the first image.
[tex]Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}[/tex]
[tex]Vout = 5 \frac{2000}{5000}[/[tex]
[tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V[/tex]
For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.
so.
[tex]R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200[/tex]
Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.
if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:
[tex]Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V[/tex]
[tex]Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V[/tex]
[tex]R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140[/tex]
[tex]R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260[/tex]
so.
[tex]V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}[/tex]
