Negative! ('scuse the pun).
A circuit consists of two wires. Switching either would produce a spark. The switch rating is (say) 200-Amps - that means, once closed, the metal parts in the switch can 'handle' that current. The same as when touching two connectors together, the switch metal will spark as the two parts touch, eventually burning away the contact(s). Even then, the switch introduces a small value of resistance in circuit, causing heat losses.
The current (at time of contact) will be in the order of thousands of amps. The capacitor 'sucks' the current according to 'little-r' (or Equivalent Series Resistance) in the order of 25 milli-Ohms. Applying Ohms' Law - Amps = Volts ÷ Resistance. Substituting: 40V (6S) ÷ 0.025 Ohms (say) = 1600 Amps (for about 20 milli-Seconds, until the capacitor(s) charge up)!
Even a large arc-welder would do well to reach this level. The figures above are representative, but give the idea.
You have to introduce a current limiting fixed resistor temporarily in series with the (+ve or -ve) lead to lower the charging current, and allow the time for the capacitor(s) to fully charge; then 'short-out' this resistor to allow the circuit to draw the required current to operate efficiently. Remember that a circuit (device) draws current relative to the applied voltage and the resistance (impedance) of the device.
Capacitors are a virtual short-circuit over the battery until charged.
This Is The (Physical) Law ...