For the circiut shown in the figure, I=0.50 A and R=12 ohms. What is the value of the emf?

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It is neither series or parallel. It is a 2 mesh circuit or a 1 node.
We know the 2 – 24 resistors are in parallel with each other and must have the same voltage (both of their ends are connected to each other). We know the voltage across the top right 24Ω must be 0.5*24Ω = 12V so the vertical 24Ω also has 12V. That means there must be 0.5A down into the vertical 24Ω resistor. This means there must be 0.5+0.5 = 1A from left to right through the 12Ω resistor. That 1A divides in half at the junction of the 2 – 24Ω resistors which makes sense since it sees 24Ω in each path. The KVL around the left loop in a clockwise direction is
+E – 12V – 12V = 0 => E = 24V
OR we could see there are 2 – 24Ω in parallel in the right loop = 12Ω. Thus E sees 12Ω+12Ω = 24Ω and there is 1A out of E so E = 1A*24Ω = 24V
OR we could solve using a single node V just above the vertical 24Ω:
(V-E)/12 + V/24 + V/24 = 0 But we know V = 0.5A*24Ω = 12V (bottom of circuit chosen to be ground) so set V = 12
(12-E)/12 + 12/24 + 12/24 = 0
12/12 + 12/24 +12/24 = E/12
12 + 6 + 6 = E = 24V

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