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

Question

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 (A)  (E1+E2/r1+r2)× R (B)  (E1+E2/r1+r2+R)× R (C)  (E1-E2/r1+r2)× R (D)  (E1-E2/r1+r2+R)× R

Tardigrade Admin · Accepted Answer

Given, two cells of emf E1 and E2 are joined in opposition such that, E1> E2. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0ca881cf5aee326a5e3820b0732e2fea-.png" /> Current, I=(E1-E2/r1+r2+R) Now, terminal potential difference across a circuit, V=I R V=((E1-E2/I1+I2+R)) R

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

$\frac{E _{1} + E _{2}}{r _{1} + r _{2}} \times R$

$\frac{E _{1} + E _{2}}{r _{1} + r _{2} + R} \times R$

$\frac{E _{1} - E _{2}}{r _{1} + r _{2}} \times R$

$\frac{E _{1} - E _{2}}{r _{1} + r _{2} + R} \times R$

Given, two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition such that, $E_{1} > E_{2}$ .

Current, $I = \frac{E _{1} - E _{2}}{r _{1} + r _{2} + R}$
Now, terminal potential difference across a circuit,
$V = I R$
$V = (\frac{E _{1} - E _{2}}{I _{1} + I _{2} + R}) R$

Q. 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

r1​+r2​E1​+E2​​×R

r1​+r2​+RE1​+E2​​×R

r1​+r2​E1​−E2​​×R

r1​+r2​+RE1​−E2​​×R