Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are R1 and R2(R2R1). If the potential difference across the source having internal resistance R2 is zero, then

Question

Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are R1 and R2(R2>R1). If the potential difference across the source having internal resistance R2 is zero, then (A) R=(R2 ×(R1+R2)/(R2-R1)) (B) R=R2-R1 (C) R=(R1 R2/(R1+R2)) (D) R=(R1 R2/(R2-R1))

Tardigrade Admin · Accepted Answer

Re q=R1+R2+R ∴ I ≡ (2 E/R1+R2+R) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f550eebe117002a84ae099f5f2d4096e-.png" /> According to the question, -(VA-VB)=E-I R2 0=E-I R2 E=I R2 E=(2 E/R1+R2+R) R2 R1+R2+R=2 R2 ∴ R=R2-R1

Q. Two sources of equal emf are connected to an external resistance $R$ . The internal resistances of the two sources are $R_{1}$ and $R_{2} (R_{2} > R_{1})$ . If the potential difference across the source having internal resistance $R_{2}$ is zero, then

$R = \frac{R _{2} \times ( R _{1} + R _{2} )}{( R _{2} - R _{1} )}$

$R = R_{2} - R_{1}$

$R = \frac{R _{1} R _{2}}{( R _{1} + R _{2} )}$

$R = \frac{R _{1} R _{2}}{( R _{2} - R _{1} )}$

$R_{e q} = R_{1} + R_{2} + R$
$∴ I \equiv \frac{2 E}{R _{1} + R _{2} + R}$

According to the question,
$- (V_{A} - V_{B}) = E - I R_{2}$
$0 = E - I R_{2}$
$E = I R_{2}$
$E = \frac{2 E}{R _{1} + R _{2} + R} R_{2}$
$R_{1} + R_{2} + R = 2 R_{2}$
$∴ R = R_{2} - R_{1}$

Q. Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are R1​ and R2​(R2​>R1​). If the potential difference across the source having internal resistance R2​ is zero, then

R=(R2​−R1​)R2​×(R1​+R2​)​

R=R2​−R1​

R=(R1​+R2​)R1​R2​​

R=(R2​−R1​)R1​R2​​