0 votes 0 votes The current $I$ through resistance $\mathrm{r}$ in the circuit shown in the figure is $\frac{-\mathrm{V}}{12 \mathrm{R}}$ $\frac{\mathrm{V}}{12 \mathrm{R}}$ $\frac{\text{V}}{6 \text{R}}$ $\frac{\mathrm{V}}{3 \mathrm{~T}}$ Others gate1998-ec + – admin asked Sep 26, 2022 • edited Dec 1, 2023 by makhdoom ghaya admin 46.4k points 32 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
