I-nE-nR-r - make -n- the subject of formula-

The correct option is A n=Ir/E IR I = nEnR+r ⇒ InR + Ir = nE ⇒ InR #160; nE = Ir ⇒ n(IR E) = Ir ⇒ n = IrE IR ...

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Byju's Answer

Standard XII

Mathematics

Definition of Functions

I=nE/nR+r ; m...

Question

$I = \frac{n E}{n R + r}$ ; make 'n' the subject of formula.

n = \frac{I r}{E - I R^{2}}

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n = \frac{I r^{2}}{E - I R}

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n = \frac{I r}{E - I R}

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none of the above

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Solution

The correct option is A $n = \frac{I r}{E - I R}$
$I = \frac{n E}{n R + r}$
$\Rightarrow I n R + I r = n E$
$\Rightarrow I n R - n E = - I r$
$\Rightarrow n (I R - E) = - I r$
$\Rightarrow n = \frac{I r}{E - I R}$

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I=nEnR+r ; make 'n' the subject of formula.

The correct option is A n=IrE−IRI=nEnR+r⇒InR+Ir=nE⇒InR−nE=−Ir⇒n(IR−E)=−Ir⇒n=IrE−IR

$I = \frac{n E}{n R + r}$ ; make 'n' the subject of formula.

The correct option is A $n = \frac{I r}{E - I R}$
$I = \frac{n E}{n R + r}$
$\Rightarrow I n R + I r = n E$
$\Rightarrow I n R - n E = - I r$
$\Rightarrow n (I R - E) = - I r$
$\Rightarrow n = \frac{I r}{E - I R}$