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These 2 circuits consist of voltage source, 2 capacitors in series and their discharging resistors.

Capacitors C1 and C2 are put in series because voltage V1 is higher than rated voltage of each capacitor. Nominal values of capacitors and resistors are the same (C1 = C2, R1 = R2). Only difference between both circuits is the link between points A and B.

schematic

simulate this circuit – Schematic created using CircuitLab

In case that the real values of R and C match their expected nominal values, both circuits should behave in the same manner, since there is no voltage difference between points A and B.

However, if, for example, capacitance of C2 drops due to wear and voltage in point A increases, there will be (initially) a voltage difference between points A and B. That higher voltage could possibly further damage the capacitor. In case of Circuit 2, current will flow through the link A-B and the circuit will eventually reach an equilibrium, where the voltage on the A-B link will be V/2 (if R1 = R2). This could (possibly) prevent the capacitor C2 from total failure due to overvoltage. Therefore, from this point of view, I would say it is better to add the A-B link, as shown in the Circuit 2.

My question is, could there be any negative effects of adding the A-B link?

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  • \$\begingroup\$ There cannot be a voltage difference between points A and B - they are connected by a short circuit. \$\endgroup\$ – Chu Sep 5 '19 at 16:22
  • \$\begingroup\$ In Circuit 1 there will be a voltage difference. In Circuit 2 of course not, but there will be current flowing through the link before the circuit reaches an equilibrium state. \$\endgroup\$ – simonov Sep 5 '19 at 16:26
  • \$\begingroup\$ If C1 = C2 and they are not leaky then voltage at point A will be 1/2 V1 = B. The short from A to B insures the voltages will be the same even if components drift. \$\endgroup\$ – Rob B. Sep 5 '19 at 17:14
  • \$\begingroup\$ meant ensures not insures \$\endgroup\$ – Rob B. Sep 5 '19 at 17:19
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It's essential to link points A and B.

Even if the capacitors have the same value, if they have different leakage currents, then under DC conditions, the potential of point A, which was initially halfway between the rails, will drift towards one or other rail, over-volting one of the capacitors.

R1 and R2 should conduct at least 10x the expected leakage current, in order to keep the voltages balanced.

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If R1 shorts out and V1 exceeds the rating for C2, then the C2 could be damaged. Same for R2 and C1. You could put a bleed resistor between A and B.

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