Four cells, each of emf E and internal resistance r, are connected in series across an external resistance R. By mistake one of the cells is connected in reverse. Then the current in the external circuit is

Question

Four cells, each of emf E and internal resistance r, are connected in series across an external resistance R. By mistake one of the cells is connected in reverse. Then the current in the external circuit is (A) (2E/4r+R) (B) (3E/4r+R) (C) (3E/3r+R) (D) (2E/3r+R)

Tardigrade Admin · Accepted Answer

Total emf of the cell =3 E-E=2 E <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a96dbc6a08edee8d08a5697c6ba45f92-.png" /> Total internal resistance =4 r ∴ Total resistance of the circuit =4 r+R So, the current in the external circuit (∵ i=(V/R)) ∴ i=(2 E/4 r+R)

Q. Four cells, each of emf $E$ and internal resistance $r$ , are connected in series across an external resistance $R$ . By mistake one of the cells is connected in reverse. Then the current in the external circuit is

$\frac{2 E}{4 r + R}$

$\frac{3 E}{4 r + R}$

$\frac{3 E}{3 r + R}$

$\frac{2 E}{3 r + R}$

Total emf of the cell $= 3 E - E = 2 E$

Total internal resistance $= 4 r$
$∴$ Total resistance of the circuit $= 4 r + R$
So, the current in the external circuit $(∵ i = \frac{V}{R})$
$∴ i = \frac{2 E}{4 r + R}$

Q. Four cells, each of emf E and internal resistance r, are connected in series across an external resistance R. By mistake one of the cells is connected in reverse. Then the current in the external circuit is

4r+R2E​

4r+R3E​

3r+R3E​

3r+R2E​