Explanation

(a) Voltage across parallel combination is same. The equivalent resistance of upper branch
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R = 4 + 6 = 10\ \Omega$
So, $\ I_{1} \times 10 = I_{2} \times 10$
Or $\ \ \ \ \ \ \ \ \ \ \ I_{1} = I_{2} = I$ (say)
$ie,$ Same current is flowing through both branches.
Now, it is given that
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ H_{4\Omega} = 10\ cal$
Or $I^{2}t\ \times 4 = 10$
Or $I^{2}t = \frac{10}{4}$
Therefore, heat produced in $10\ \Omega$ resistance is
$\ \ \ \ \ \ \ \ \ \ \ \ \ H_{10\Omega} = 10 \times I^{2}t = \frac{10 \times 10}{4} = 25\ cal$