No 38 tolong ya.. Makasih

Hambatan di rangkaian paralel:

[tex] = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \\ \\ = \frac{3 + 2 + 1}{6} \\ \\ = \frac{6}{6} \\ \\ = 1[/tex]

Hambatan total: Hambatan di rangkaian seri + Hambatan di rangkaian paralel
= 1 + 1
= 2 Ω

Beda Potensial (V) = I x R
= 1,5 x 2
= 3 V

Jawabannya C.