Question

The output voltage $V_0$ when the jockey of the potentiometer is exactly at the midpoint, would be

• A. -4 V
• B. -2.30 V
• C. -2.83 V
• D. -1.72 V

Answer 1

The correct option is B -2.30 V

$V_0=\frac{-30}{20}\times V_A[\because V_0=\frac{-R_F{V}_i}{R_3}]$

$V_0=-1.5V_A$

$\text{To calculate}{V}_A,$

$\text{Apply KCL at node B,}$

$\frac{10-V_B}{25k}=\frac{V_B-0}{25k}+\frac{V_B-V_A}{10k}$

$\frac{10}{25}+\frac{V_A}{10}=V_B[\frac{1}{25}+\frac{1}{25}+\frac{1}{10}]$

$\frac{10}{25}+\frac{V_A}{10}=V_B\left[\frac{2+2+5}{50}\right]$

$\frac{20+5V_A}{50}=\frac{9V_B}{50}$

$V_B=\frac{1}{9}[20+5V_A]$

$\text{Apply KCL at Node A,}$

$\frac{V_A-V_B}{10k}+\frac{V_A}{20k}+\frac{V_A-V_0}{50k}=0$

$V_A[\frac{1}{10}+\frac{1}{20}+\frac{1}{50}]=\frac{V_B}{10}+\frac{V_0}{50}$

$V_A\times \left[\frac{17}{100}\right]=\frac{1}{9}[20+5V_A]\times \frac{1}{10}-\frac{1.5V_A}{50}$

$V_A\times \left[\frac{17}{100}\right]=\frac{20}{90}+\frac{5V_A}{90}-\frac{1.5V_A}{50}$

$V_A\times \frac{17}{100}-\frac{5V_A}{90}+\frac{1.5V_A}{50}=\frac{20}{90}$

$V_A\left[\frac{153-50+27}{900}\right]=\frac{20}{90}$


$V_A\times \frac{130}{900}=\frac{20}{90}$

$V_A=1.538$

$V_0=-1.5V_A=-2.307V$