Boost converter working

Boost converter is a DC to DC converter
Output voltage is greater than input voltage
Hence it is a step up converter
During switch on inductor will get charged at that time load get supply from capacitor.
Capacitor voltage will not go to source side due to the presence of diode. Diode will be reverse biased
During switch off inductor will discharge so both voltage source and inductor will act as a source and will deliver power to the load. At this time capacitor will get charged
Capacitor is parallel to load therefore both will have same voltage because element connected in parallel have same voltage.

Apply KVL to get inductor voltage
Apply KCL to get capacitor current

Vo is greater than Vs

output voltage equation is obtained from volt-sec balance equation of inductor.
Average voltage across an inductor in DC- steady state will be zero.

output current equation is obtained from Amp-sec balance equation of capacitor
Average current in capacitor will be zero.

[sammyfallatah]

what if my capacitor was polarized to reduce ESR would it's polarity change like the inductor?