Current measured by the ammeter (A) in the reported circuit when no...

This is a Wheatstone circuit. For the current to be zero in $10Ω$ the Wheatstone circuit should be balanced. Therefore,

$\dfrac{R}{3} = \dfrac{4}{6} ⇒ R = 2Ω$

Now, the equivalent resistance across the battery will be,

${R}_{eq} = \dfrac{\left(2 + 4\right) \times \left(3 + 6\right)}{\left(2 + 4\right) + \left(3 + 6\right)} = 3 . 6Ω$

Therefore, the current measured by the ammeter is $i = \dfrac{V}{R} = \dfrac{36}{3 . 6} = 10A$.