In the circuit shown in the figure, the rheostat has a resistance...

The equivalent resistance of the parallel combination of $10\text{ }\Omega$ and $20\text{ }\Omega$ resistors is:

$R_p = \dfrac{10 \times 20}{10 + 20} = \dfrac{200}{30} = \dfrac{20}{3}\text{ }\Omega$

The total resistance of the circuit is the sum of $R_p$ and the resistance of the rheostat $R_v$ included in the circuit:

$R_{eq} = R_p + R_v = \dfrac{20}{3} + R_v$

The current in the circuit is given by Ohm's law:

$I = \dfrac{V}{R_{eq}} = \dfrac{5.5}{\dfrac{20}{3} + R_v}$

For maximum current, the rheostat resistance should be minimum, i.e., $R_v = 0\text{ }\Omega$:

I_\max = \dfrac{5.5}{\dfrac{20}{3} + 0} = \dfrac{5.5 \times 3}{20} = \dfrac{16.5}{20} = 0.825\text{ A} \approx 0.83\text{ A}

For minimum current, the rheostat resistance should be maximum, i.e., $R_v = 30\text{ }\Omega$:

I_\min = \dfrac{5.5}{\dfrac{20}{3} + 30} = \dfrac{5.5}{\dfrac{110}{3}} = \dfrac{5.5 \times 3}{110} = \dfrac{16.5}{110} = 0.15\text{ A}

Thus, the minimum and maximum currents are $0.15\text{ A}$ and $0.83\text{ A}$ respectively.