Two cells of unequal emfs, `varepsilon_(1)` and `varepsilon_(2)` and internal resistances `r_(1)` and `r_(2)` are joined as shown. `V_(A)` and `V_(B)` are the potentials at `A` and `B` respectively.

Two cells of unequal emfs epsi_(1)" and "epsi_(2) , and internal resistances r_(1)" and "r_(2) are joined as shown. V_(A)" and "V_(B) are the potentials as A and B respectively.

Two cells of unequal emfs epsilon_1 and epsilon_2 , and internal resistance r_1 and r_2 are joined as shown. V_A and V_B are the potential at A and B respectively. (i) One cell will continuously supply energy to the other (ii) The potential difference across both the cells will be equal (iii) The potential difference across one cell with be greater than its emf. (iv) V_A - V_B = (epsilon_1 r_1 + epsilon_2 r_2)/(r_1 + r_2) .

Assertion:- To cells of unequal emf E_(1) and E_(2) having internal resistances r_(1) and r_(2) are connected as shown in figure. Then the potential difference across any cell cannot be zero. Reason:- If two cells having nonzero internal resistance and unequal emf are connected across each other as shown, then the current in the circuit cannot be zero.

Two cells of emfs epsilon_(1)and epsilon_(2) and internal resistances r_(1)and r_(2) respectively are connected in parallel . Obtain expressions for the equivalent (i) resistance and , (ii) emf of the combination.

Two cells of emfs epsilon_(1)and epsilon_(2) and internal resistances r_(1)and r_(2) respectively are connected in parallel . Obtain expressions for the equivalent (i) resistance and , (ii) emf of the combination.

Two capacitors C_(1) and C_(2) in a circuit are joined as shown. If V_(A)=V_(1),V_(B)=V_(2) , the potential of point P is

Two cells of emf E_(1) and E_(2) have internal resistance r_(1) and r_(2) . Deduce an expression for equivalent emf of their parallel combination.

Two cells of emf 3V and 5V and internal resistance r_(1) and r_(2) respectively are in series with an external resistance R. if the p.d. across 1 st cell is zero, then R is