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Dec 30, 2021 at 3:26 vote accept across
Dec 30, 2021 at 3:26 comment added across By small change in \$V-E\$, I mean when \$V-E\$ is very small, a multiple of it: \$a*(V-E)\$ is also very small. Thanks again I totally get it XD
Dec 30, 2021 at 3:12 comment added across Ahh I get it now thanks to you :) \$I = \dfrac{V-E}{R}\$. If \$R\$ is small enough, a large change in load/current can be balanced by a small change in \$V-E\$
Dec 30, 2021 at 3:10 comment added D.A.S. Any followup questions?
Dec 30, 2021 at 2:59 history answered D.A.S. CC BY-SA 4.0