Answer
The emf of the battery is $48.3~V$
Work Step by Step
$i_5 = i_6 = 1.40~A$
The potential difference across $R_5$ and $R_6$:
$\Delta V = (1.40~A)(8.00~\Omega)+(1.40~A)(4.00~\Omega) = 16.8~V$
This is the same potential difference across $R_4$. The current through $R_4$:
$i_4 = \frac{16.8~V}{16.0~\Omega} = 1.05~A$
The total current through $R_2$ is $1.40~A+1.05~A = 2.45~A$
The potential difference across $R_2$ is $(2.45~A)(2.00~\Omega) = 4.90~V$
The total potential difference across $R_2$ and $R_4$ is $4.90~V+16.8~V = 21.7~V$
This is the same potential difference across $R_3$
The current through $R_3$ is $\frac{21.7~V}{2.00~\Omega} = 10.85~A$
The total current through $R_1$ is $10.85~A+2.45~A = 13.3~A$
The potential difference across $R_1$ is $(13.3~A)(2.00~\Omega) = 26.6~V$
The total potential difference across $R_1$ and $R_3$ is $26.6~V+21.7~V = 48.3~V$
This is equal to the emf of the battery.
The emf of the battery is $48.3~V$
