N identical cells, each of emf E and internal resistance r are joined in series. Out of N cells, n cells are wrongly connected i.e., their terminals are connected in reverse of that required for series connection (n

Question

N identical cells, each of emf E and internal resistance r are joined in series. Out of N cells, n cells are wrongly connected i.e., their terminals are connected in reverse of that required for series connection (n < (N/2)) . Let E0 be the emf of resulting battery and r0 be its internal resistance. Then (A) E0=(N-n) E, r0=(N-n) r (B) E0=(N-2 n) E, r0=(N-2 n) r (C) E0=(N-2 n) E, r0=N r (D) E0=(N-n) E, r0=N r

Tardigrade Admin · Accepted Answer

For every cell that is wrongly connected, the emf decreases by 2 E. However, internal resistance does not depend on the direction and therefore remains same for all cells.

$E_{0} = (N - n) E, r_{0} = (N - n) r$

$E_{0} = (N - 2 n) E, r_{0} = (N - 2 n) r$

$E_{0} = (N - 2 n) E, r_{0} = N r$

$E_{0} = (N - n) E, r_{0} = N r$

For every cell that is wrongly connected, the emf decreases by $2 E$ . However, internal resistance does not depend on the direction and therefore remains same for all cells.

E0​=(N−n)E,r0​=(N−n)r

E0​=(N−2n)E,r0​=(N−2n)r

E0​=(N−2n)E,r0​=Nr

E0​=(N−n)E,r0​=Nr

For every cell that is wrongly connected, the emf decreases by 2E. However, internal resistance does not depend on the direction and therefore remains same for all cells.

$E_{0} = (N - n) E, r_{0} = (N - n) r$

$E_{0} = (N - 2 n) E, r_{0} = (N - 2 n) r$

$E_{0} = (N - 2 n) E, r_{0} = N r$

$E_{0} = (N - n) E, r_{0} = N r$

For every cell that is wrongly connected, the emf decreases by $2 E$ . However, internal resistance does not depend on the direction and therefore remains same for all cells.