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

(A) $R=R_{2} \times\left(R_{1}+R_{2}\right) / R_{2}-R_{1}$

(C) $R=R_{1} R_{2} /\left(R_{1}+R_{2}\right)$

(D) $R=R_{1} R_{2} /\left(R_{2}-R_{1}\right)$

JEE Main · 2005 · 08 May 2005 · Offline · Current Electricity

Two sources of equal emf are connected to an external resistance

R

. The internal resistance 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

Answer: (B)

With the two equal emfs

E

driving current through the external

R

, the current is

I = \frac{2 E}{R + R _{1} + R _{2}}

. The terminal voltage of the cell whose internal resistance is

R_{2}

E - I R_{2}

, and setting it to zero gives

E = I R_{2}

. Substituting

I

E = \frac{2 E R _{2}}{R + R _{1} + R _{2}}

, so

R + R_{1} + R_{2} = 2 R_{2}

and

R = R_{2} - R_{1}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option