If I'm understand you correctly then yes adding a diode in parallel will keep the electricity running in the series circuit and keep everything lit. Calculating the resistance in this case should be the same as previous. You would act as if the bulb that was not working was just not part of the series circuit.
@@tradestutor1191 ..can you proof it in a demonstration? ..I can't find anything similar to that in any RU-vid vid.. demonstrate using a 2 x 5 mm LED in series with at least 2 bypass diode in parrellel with both LED and it's values..
@@surenbono6063 Sorry: I'm not going to have time to do that anytime soon. Maybe check out a channel called "The Electric Academy". It goes through a lot of stuff and maybe some of it will help you out.
