How do you solve for n in # I = (nE)/(nr+R)#?

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Answer 1 · 2016-04-27T22:33:52

#n = (RI)/(E-rI)#

Given:

#I = (nE)/(nr+R)#

Transpose and multiply both sides by #(nr+R)# to get:

#nE = (nr+R)I = nrI+RI#

Transpose and subtract #nrI# from both sides to get:

#RI = nE - nrI = n(E-rI)#

Transpose and divide both sides by #(E-rI)# to get:

#n = (RI)/(E-rI)#