Answer
The ammeter reads $~~0.45~A$
Note that this is the same value that was found in part (a)
Work Step by Step
Suppose the ammeter and the battery are interchanged.
We can find the equivalent resistance of $R_2$ and $R_1$:
$\frac{1}{R_{12}} = \frac{1}{4.0~\Omega}+ \frac{1}{2.0~\Omega}$
$R_{23} = 1.33~\Omega$
Then the equivalent resistance in the circuit is $1.33~\Omega+ 6.0~\Omega = 7.33~\Omega$
We can find the total current in the circuit:
$i = \frac{\mathscr{E}}{7.33~\Omega} = \frac{5.0~V}{7.33~\Omega} = 0.682~A$
We can find the potential difference across $R_3$:
$\Delta V = (0.682~A)(6.0~\Omega) = 4.1~V$
Then the potential difference across $R_1$ is $~~5.0~V-4.1~V = 0.90~V$
We can find the current through $R_1$:
$i_1 = \frac{0.90~V}{2.0~\Omega} = 0.45~A$
Since this is the same current through the ammeter, the ammeter reads $~~0.45~A$
Note that this is the same value that was found in part (a)
