Simplify the complex fraction.
-$I = \frac { n E } { n r + R }$ for $n$
A) $n = \frac { I R } { E - I r }$
B) $\mathrm { n } = \frac { \mathrm { I } ( \mathrm { nr } + \mathrm { R } ) } { \mathrm { E } }$
C) $n = I R - E + I r$
D) $\mathrm { n } = \frac { \mathrm { IR } } { \mathrm { Ir } - \mathrm { E } }$

Question

Quizplus · Accepted Answer

To solve for $n$, we start with the given equation $I = \frac { n E } { n r + R }$. Multiplying both sides by $n r + R$ gives us $I (n r + R) = n E$. Expanding the left side, we get $Inr + IR = nE$. Rearranging to solve for $n$, we isolate $n$ on one side: $nE - Inr = IR$. Factoring $n$ out of the left side gives us $n(E - Ir) = IR$. Finally, dividing both sides by $E - Ir$ gives us $n = \frac { I R } { E - I r }$.

Simplify the Complex Fraction I=nEnr+RI = \frac { n E } { n r + R }I=nr+RnE​

Simplify the Complex Fraction $I = \frac { n E } { n r + R }$