Consider the circuit shown in the figure. The value of current ' $I$ ' is
- $-\frac{7}{18} \mathrm{~A}$
- $5 \mathrm{~A}$
- $3 \mathrm{~A}$
- $-3 \mathrm{~A}$
Solution
Current from B to A, $\mathrm{I}_1=\frac{6-(-8)}{28}=0.5 \mathrm{~A}$
Current from $\mathrm{C}$ to $\mathrm{B}=\mathrm{I}_2=\frac{12-6}{54}=\frac{1}{9} \mathrm{~A}$
Hence, $\mathrm{I}=\mathrm{I}_2-\mathrm{I}_1=\frac{1}{9}-\frac{1}{2}=-\frac{7}{18} \mathrm{~A}$
.Asked in: MHT CET 2021 (20 Sep Shift 2)