a) The combined resistance of the above circuit. Step 1: Calculate the equivalent resistance of the parallel combination of the $4\,\Omega$ and $6\,\Omega$ resistors. $$ \frac{1}{R_p} = \frac{1}{4\,\Omega} + \frac{1}{6\,\Omega} $$ $$ \frac{1}{R_p} = \frac{3}{12\,\Omega} + \frac{2}{12\,\Omega} $$ $$ \frac{1}{R_p} = \frac{5}{12\,\Omega} $$ $$ R_p = \frac{12}{5}\,\Omega = 2.4\,\Omega $$ Step 2: Calculate the total combined resistance ($R_{total}$) by adding the series resistor ($2\,\Omega$) to the parallel equivalent resistance ($R_p$). $$ R_{total} = 2\,\Omega + R_p $$ $$ R_{total} = 2\,\Omega + 2.4\,\Omega $$ $$ R_{total} = 4.4\,\Omega $$ The combined resistance of the circuit is $\boxed{\text{4.4 \Omega}}$. b) The current flowing through the ammeter. Step 1: Use Ohm's Law ($V = I \cdot R$) to find the total current ($I_{total}$) flowing from the voltage source, which is measured by the ammeter. $$ I_{total} = \frac{V_{source}}{R_{total}} $$ $$ I_{total} = \frac{12\,V}{4.4\,\Omega} $$ $$ I_{total} = \frac{120}{44}\,A = \frac{30}{11}\,A $$ $$ I_{total} \approx 2.73\,A $$ The current flowing through the ammeter is $\boxed{\text{2.73 A}}$.