Answer
$V_3 = 6V $ and $I_3 = 2A$
$V_{16} = 6V $ and $ I_{16} = 0.375A $
$V_{48} = 6V $ and $ I_{48} = 0.125A$
$V_{4} = 6V $ and $ I_{4} = 1.5A$
Work Step by Step
$16\Omega$, $48\Omega$ and $4\Omega$ are in parallel with each other. this combination is in series with$3\Omega$. thus
$R_{eq} =6\Omega$
now $I = \frac{V}{R_{eq}} =\frac{12}{6}A = 2A$
current through $3\Omega$ is $2A$
potential drop across $3\Omega$
$V_3 = IR =2*3 V = 6V$
potential drop across $16\Omega$, $48\Omega$ and $4\Omega$ is same and is equal to
$V = 12V - 6V = 6 V$
now
$V_3 = 6V $ and $I_3 = 2A$
$V_{16} = 6V $ and $ I_{16} = \frac{6}{16}A =0.375A $
$V_{48} = 6V $ and $ I_{48} = \frac{6}{48}A = 0.125A$
$V_{4} = 6V $ and $ I_{4} = \frac{6}{4}A = 1.5A$
