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enter image description here

If I were to add a second capacitor to this circuit of roughly the same capacitance (120uf). Would this circuit charge both up completely, and would the LED still accurately depict if the combined two capacitors are charged. Does adding another capacitor not work or cause the circuit to not work?

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  • \$\begingroup\$ "If I were to add a second capacitor to this circuit of roughly the same capacitance (120uf)". Where would you put it? \$\endgroup\$ Jan 2 '18 at 0:33
  • \$\begingroup\$ In line with the other one to make it like one bigger capacitor. \$\endgroup\$
    – techset
    Jan 2 '18 at 0:52
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    \$\begingroup\$ You have confused "in line" with "across". \$\endgroup\$
    – gbarry
    Jan 2 '18 at 1:51
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A second identical capacitor in parallel with the first would double the energy available to the flash, resulting in brighter flashes, longer charge-up time and fewer flashes before the tube fails.

It's not an LED it's a neon, but yeah it responds to the voltage in the capacitors, so it will still indicate that the circuit is ready to fire.

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  • \$\begingroup\$ it seems to take less time for the light/led/neon to become a solid red color meaning it's ready. Shouldn't it take longer if anything, that is what i don't understand. \$\endgroup\$
    – techset
    Jan 2 '18 at 1:39
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    \$\begingroup\$ Taking your other comment literally, you just halved the capacitance when you put them in series. \$\endgroup\$
    – gbarry
    Jan 2 '18 at 1:52
  • \$\begingroup\$ yeah if you connect them in series, time would be reduced and the tube would last longer. \$\endgroup\$
    – Jasen
    Jan 2 '18 at 3:20

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