Question

How do you solve for $n$ in $I = \dfrac{{nE}}{{nr + R}}$?

Answer 1

Hint:In the given equation, $I$ is given in terms of $E$, $n$ , $r$ and $R$. Solving for $n$ here means rearranging the given equation and writing $n$ in terms of all other variables in the equation. For this, $n$ is to be shifted to LHS and other variables to RHS such that $n$ becomes the dependent variable, the value of which depends on the other independent variables, i.e.,$E$, $I$ , $r$ and $R$.

Complete step by step solution:
Given equation is $I = \dfrac{{nE}}{{nr + R}}$
Since $n$ is on the RHS, we first interchange the terms of LHS and RHS. We get,
$\dfrac{{nE}}{{nr + R}} = I$
Now, we multiply both sides by\[(nr + R)\] to get:
$ \Rightarrow \dfrac{{nE}}{{(nr + R)}} \times (nr + R) = I \times (nr + R)$
We find that the expression$(nr + R)$appears both in the numerator and the denominator on the LHS, which would yield the value of 1. Thus, we get:
$ \Rightarrow nE = I \times (nr + R)$
Applying distributive property on the RHS, we get:
$ \Rightarrow nE = Inr + IR$
Subtracting the expression $Inr$ from both sides, we get:
$
\Rightarrow nE - Inr = Inr + IR - Inr \\
\Rightarrow nE - Inr = IR \\
$
Taking \[n\] common from both the terms in the LHS, we get:
$ \Rightarrow n(E - Ir) = IR$
Now, we divide both sides by $(E - Ir)$. Here we have to realise that the expression $(E - Ir)$is a non- zero number because dividing a number by zero gives an undefined result.
Thus, we get:
\[ \Rightarrow \dfrac{{n(E - Ir)}}{{(E - Ir)}} = \dfrac{{IR}}{{(E - Ir)}}\]
We find that the expression $(E - Ir)$ appears both in the numerator and the denominator on the LHS, which would yield the value of 1. Thus, we get:
\[ \Rightarrow n = \dfrac{{IR}}{{(E - Ir)}}\]
From this equation, we get $n$ in terms of the variables $E$ , $I$ , $r$ and $R$. Knowing the value of $$ , $I$ , $r$ and $R$, we can solve the above equation to get the value of $n$.

Additional Information:
The given equation is the formula to calculate current $I$ in an electric circuit having electric potential $E$ with $n$ number of resistors with resistance $R$ connected in parallel to each other which in turn is connected in series with a resistor of resistance $r$.

Note: Here, we have assumed that$(E - Ir) \ne 0$. If$(E - Ir) = 0$, the value of $n$ will become undefined as any expression or number divided by $0$is undefined. Writing $n$ in terms of other variables means $n$ is the dependent variable which is dependent on independent variables $E$ ,$I$ , $r$ and $R$.