Two cells, having the same E.M.F. are connected in series through an external resistance R . Cells have internal resistances r1 and r2 (r1 r2) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is

Question

Two cells, having the same E.M.F. are connected in series through an external resistance R . Cells have internal resistances r1 and r2 (r1 > r2) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is (A) (r1 + r2/2) (B) (r1 - r2/2) (C) r1+r2 (D) r1-r2

Tardigrade Admin · Accepted Answer

Current in the circuit =(E + E/r1 + r2 + R)=(2 E/r1 + r2 + R) The potential difference across the first cell V1=E-ir1 ⇒ V1=E-(2 E × r1/(r1 + r2) + R) Now, E-(2 E r1/(r1 + r2) + R)=0 ⇒ E=(2 E r1/(r1 + r2) + R)⇒ 2r1=r1+r2+R ⇒ R=r1-r2

Q. Two cells, having the same E.M.F. are connected in series through an external resistance $R$ . Cells have internal resistances $r_{1}$ and $r_{2}$ $(r_{1} > r_{2})$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is

$\frac{r _{1} + r _{2}}{2}$

$\frac{r _{1} - r _{2}}{2}$

$r_{1} + r_{2}$

$r_{1} - r_{2}$

Q. Two cells, having the same E.M.F. are connected in series through an external resistance R . Cells have internal resistances r1​ and r2​ (r1​>r2​) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is

2r1​+r2​​

2r1​−r2​​

r1​+r2​

r1​−r2​

Current in the circuit =r1​+r2​+RE+E​=r1​+r2​+R2E​ The potential difference across the first cell V1​=E−ir1​ ⇒V1​=E−(r1​+r2​)+R2E×r1​​ Now, E−(r1​+r2​)+R2Er1​​=0 ⇒E=(r1​+r2​)+R2Er1​​⇒2r1​=r1​+r2​+R ⇒R=r1​−r2​

Q. Two cells, having the same E.M.F. are connected in series through an external resistance $R$ . Cells have internal resistances $r_{1}$ and $r_{2}$ $(r_{1} > r_{2})$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is

$\frac{r _{1} + r _{2}}{2}$

$\frac{r _{1} - r _{2}}{2}$

$r_{1} + r_{2}$

$r_{1} - r_{2}$