 Transformer circuit
It lowers the AC 100 V to AC 4 V with the transformer. As for the circuit this time, the consumption electric power of the thermometer is small. So, it is enough if being 20 mA as the electric current on the side AC 4 V.
Rectification circuit
Rectifying the full wave by the diode bridge and changing the alternating current into the direct current. The voltage of the direct current becomes rather higher than the alternating voltage when rectifying. The display of the alternating voltage is using the RMS(Root Mean Squared value). This is  of the maximum voltage (the peak voltage). When rectifying the alternating voltage, when there is not a load, the voltage which is near the peak voltage with the AC voltage approximately becomes the DC voltage. Therefore, the rectified DC voltage becomes the value which is higher than the AC voltage display. It became +5.23 V this time.
Because the output voltage of the transformer is 4.3Vrms at the circuit this time, the peak voltage is 6.06V of 1.41 times. Because the voltage drop of the silicon diode of the rectifier is about 0.7 V, the output voltage of the rectifier is 6.06V - 0.7V = 5.36 V. It is the value which is near the measurement value.
Because the ripple is contained in the rectified voltage, it makes the clean DC voltage with the capacitor. Discharging by the electric charge which was stored up at the capacitor and direct current voltage's being maintained when the alternating voltage becomes 0 V.
Supplement.
The voltage, the electric current of the alternating current are not shown by the peak value and are shown by the RMS value.
Why?
Because it is convenient to calculate the electric power. The electric power when making the resistor the load is calculated by the following formula.
Electric power(P) = Vrms x Irms
Light-Emitting Diode(LED) circuit
This is usual LED lighting-up circuit. You decide the value of the resistor R1 for the electric current which flows through the LED to become about 10-20 mA. The approximately necessary resistance value can be calculated, supposing that the voltage of the LED is about 2 V.
Because the power supply this time is 5.2V, the voltage which is applied to the resistor R1 is 5.2V - 2V = 3.2V. The resistance value to apply the electric current of 10mA(0.01A) to this resistor is R1 = 3.2V / 0.01A = 320 ohm.
In case of the 12-V power supply, it is (12V-2V) / 0.01A = 1K ohm.
Resistor for the voltage drop
The forward voltage of the LED which was used this time is about 1.65 V. The power supply of the thermometer is 1.5 V. Therefore, it lowers the voltage using the resistor. I make the ratio of R2 and R3 1:10 to lower about 0.15 V(10%) of the voltage. It should make the truth 1:9. However, for the reason for the odd resistance value and so strict voltage adjustment isn't necessary, I did to R2 = 330 ohm¤ R3 = 3.3K ohm. Because the thermometer was connected with R3 in parallel, it became the exactly good value.
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