Two cells with the same e.m.f. E and different internal resistances r1 and r2 are connected in series to an external resistance R. The value of R so that the potential difference across the first cell be zero is

Question

Two cells with the same e.m.f. E and different internal resistances r1 and r2 are connected in series to an external resistance R. The value of R so that the potential difference across the first cell be zero is (A) √r1r2 (B) r1+r2 (C) r1-r2 (D) (r1+r2/2)

Tardigrade Admin · Accepted Answer

There are two batteries with emf E each and the internal resistances r1 and r2 respectively. Hence we have I(R+r1+r2)=2 E thus, I=(2 E/R+r1+r2) Now the potential difference across the first cell would be equal to V=E-I r1 . From the question, V=0, hence, E=I r1=(2 E r1/R+r1+r2), thus, R+r1+r2=2 r1, hence R=r1-r2.

Q. Two cells with the same e.m.f. $E$ and different internal resistances $r_{1}$ and $r_{2}$ are connected in series to an external resistance $R$ . The value of $R$ so that the potential difference across the first cell be zero is

$r_{1} r_{2}$

$r_{1} + r_{2}$

$r_{1} - r_{2}$

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

Q. Two cells with the same e.m.f. E and different internal resistances r1​ and r2​ are connected in series to an external resistance R. The value of R so that the potential difference across the first cell be zero is

r1​r2​​

r1​+r2​

r1​−r2​

2r1​+r2​​

Q. Two cells with the same e.m.f. $E$ and different internal resistances $r_{1}$ and $r_{2}$ are connected in series to an external resistance $R$ . The value of $R$ so that the potential difference across the first cell be zero is

$r_{1} r_{2}$

$r_{1} + r_{2}$

$r_{1} - r_{2}$

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