Answer
1$\frac{43}{48}$ A
Work Step by Step
Let the three currents are $\frac{1}{16}$ A , $\frac{1}{12}$ A and 1$\frac{3}{4}$A
total current = $\frac{1}{16}$ A+ $\frac{1}{12}$ A + 1$\frac{3}{4}$A
= $\frac{1}{16}$ + $\frac{1}{12}$ + $\frac{7}{4}$
=$\frac{1}{16}\times$ $\frac{3}{3}$ + $ \frac{1}{12} \times$ $\frac{4}{4}$ +$ \frac{7}{4} \times$ $\frac{12}{12}$
=$\frac{3}{48}$ + $\frac{4}{48}$ `+ $\frac{84}{48}$
=$\frac{91}{48}$
=1$\frac{43}{48}$ A
