0 votes 0 votes In the $\text{op-amp}$ circuit given in the figure, the load current $i_{\text{L}}$ is $-\frac{v_{s}}{R_{2}}$ $\frac{v_{s}}{R_{2}}$ $-\frac{v_{s}}{R_{\text{L}}}$ $\frac{v_{s}}{R_{1}}$ Others gate2004-ec + – admin asked Sep 25, 2022 • edited Nov 30, 2023 by makhdoom ghaya admin 46.4k points 30 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
