The art of combining R and C components

(Joseph-Henri Lévy)
During a restoration, it sometimes happens that you don't have the right resistor or the right capacitor on hand. You can make up for these shortages with odds and ends. Here are easily handled cases.

• Series assembly: The resulting resistance is the sum of the resistances (therefore higher than each resistance taken separately)
  
• Parallel assembly: The resulting resistance is smaller than the smallest of the resistances in the assembly.
  

In the cases mentioned above, the total power is distributed among each element of the assembly:

• Proportionally to R in series circuits
• Inversely proportional to R in parallel circuits

Warning! In the case of series and parallel combinations, everything becomes more complicated.

• Series assembly: the resulting capacitance is smaller than the smallest of the capacitances in the assembly. This configuration is used for voltage adjustment.
• Parallel assembly: the resulting capacitance is the sum of the capacitances (therefore higher than each capacitance taken separately).
  

Putting capacitors in series allows distributing the main voltage across the terminals of each element. Warning! Resistors in parallel with the terminals of the elements are mandatory. They ensure a uniform division of the voltage U.

Example: We need an 8µF electrolytic capacitor insulated for 800V. Knowing that a classic electrolytic capacitor has a precision of 20% (those equipping our vintage radios could be imprecise by 50%), an 8µF model can have any value between 8 - 1.6 = 6.4 µF and 8 + 1.6 = 9.6 µF. We have 22 µF/350V capacitors. Putting three in series gives a resulting capacitance of: 22/3 = 7.3 µF withstanding 3 x 350V or 1050V.

Example: We need a 20µF electrolytic capacitor insulated for 350V. Knowing that a classic capacitor has a precision of 20%, a 20µF model can have any value between 20 - 4 = 16 µF and 20 + 4 = 24 µF. We have 2 capacitors of 10 µF/350V. Putting them in parallel gives a resulting capacitance of 20 µF withstanding 350V.

Example.
We are missing a 15K resistor. Simply connect a 10K and a 4.7K in series; which gives 14.7K.

Example.
We are missing a 470K resistor. Simply put two 1M resistors in parallel ( which gives 500K). Example.
We need a 470 Ω resistor withstanding 1W. We only have ¼ W series available. This watt must be distributed over

• Either 4 equal resistors of 120 Ω in series (i.e., 480 Ω)
• Or 4 equal resistors of 2200 Ω in parallel (i.e., 550 Ω)