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Non-Ideal Battery

Two cells of ...

Question

Two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition (such that $E_{1} > E_{2}$ ). If r1 and r2 be the internal resistance and R be the external resistance, then the terminal potential difference is

$\frac{E_{1} - E_{2}}{r_{1} + r_{2}} \times R$

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$\frac{E_{1} + E_{2}}{r_{1} + r_{2}} \times R$

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$\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R$

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$\frac{E_{1} + E_{2}}{r_{1} + r_{2} + R} \times R$

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Solution

The correct option is C
$\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R$

$I = \frac{Net emf}{Total resistance} = \frac{E_{1} - E_{2}}{R + r_{1} + r_{2}} V = I R \Rightarrow V = \frac{(E_{1} - E_{2}) R}{R + r_{1} + r_{2}}$

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Two cells of emf E1 and E2 are joined in opposition (such that E1>E2). If r1 and r2 be the internal resistance and R be the external resistance, then the terminal potential difference is

The correct option is C E1−E2r1+r2+R×R I=Net emfTotal resistance=E1−E2R+r1+r2V=IR⇒V=(E1−E2)RR+r1+r2

Two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition (such that $E_{1} > E_{2}$ ). If r1 and r2 be the internal resistance and R be the external resistance, then the terminal potential difference is

The correct option is C
$\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R$

$I = \frac{Net emf}{Total resistance} = \frac{E_{1} - E_{2}}{R + r_{1} + r_{2}} V = I R \Rightarrow V = \frac{(E_{1} - E_{2}) R}{R + r_{1} + r_{2}}$