## Elementary Technical Mathematics

1$\frac{43}{48}$ A
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