An ideal battery drives a current of 5 A through a resistor. After...

Let the voltage of the ideal battery be $V$ and the resistance of the first resistor be $R$.

Initially, the current is $5 \text{ A}$, so:

$I_1 = \dfrac{V}{R} = 5 \Rightarrow V = 5R$

When a $10 \Omega$ resistor is connected in parallel, the equivalent resistance becomes:

$R_{eq} = \dfrac{10R}{10 + R}$

The new current is $6 \text{ A}$, so:

$I_2 = \dfrac{V}{R_{eq}} = 6$

Substituting $V = 5R$ and $R_{eq}$ into the equation:

$\dfrac{5R}{\dfrac{10R}{10 + R}} = 6$

$\dfrac{5(10 + R)}{10} = 6$

$\dfrac{10 + R}{2} = 6$

$10 + R = 12$

$R = 2 \Omega$