Power Supply Issue, Voltage Drop Under Load
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Voltage divider open-circuit. Figure 2. Micro attached in parallel.


Figure 1 shows the voltage divider when tested unloaded. All appears well and 3.3 V is measured on the divider terminals.
Figure 2 shows the voltage divider loaded by the micro. The micro is represented by a 33 Ω resistor which would pass 100 mA if connected to a stable 3.3 V PSU.
The combination of R4 and R5 in parallel is about 30 Ω (28.7 Ω).
Now our potential divider mid-point will give \$ 4.8 \frac {30}{100 + 30} = 4.8 \cdot 0.23 = 1.1 V \$


The moral of the story is that a potential divider will not give you a stable voltage under changing load conditions such as a micro whose current requirements will vary with time. You need to use a voltage regulator.