Answer
$ I_5 = 1.6A$ and $V_5 = 8V$
$ I_{10} = 1.6A$ and $V_{10} = 16V$
$I_6 = 2.4 A$ and $ V_6 = 14.4V$
$I_4 = 2.4 A$ and $ V_4 = 9.6V$
Work Step by Step
$6\Omega$ is in series with $4\Omega$ and $5\Omega $ is in series with$10 \Omega $ and both these combinations are in parallel with each other. Thus
$R_{eq} =6\Omega$
so $I= \frac{V}{R_{eq}} =\frac{24}{6}A = 4A$
let $I_1 $ be current through $5\Omega$ and $10\Omega$
and $I_2$ be current through $6\Omega $ and $4\Omega$
$ I_1 = \frac{24}{5+ 10}A = 1.6A$
$I_2 = \frac{24}{6+4}A = 2.4$
Now,
$ I_5 = 1.6A$ and $V_5 = 1.6*5 V = 8V$
$ I_{10} = 1.6A$ and $V_{10} = 1.6*10 V =16V$
$I_6 = 2.4 A$ and $ V_6 = 2.4* 6V = 14.4V$
$I_4 = 2.4 A$ and $ V_4 = 2.4* 4V =9.6V$
