Two resistors with resistances `10 Omega and 15 Omega` are to be connected to a battery of `emf 12 V` so as to obtain :
(i) minimum current
(ii) maximum current. How will you connect the resistances in each case ? Calculate the strength of the total current in the circuit in the two cases.

Answer

`2 A`

Explanation

To obtain minimum current, the two resistors should be connected in series so that the equivalent resistance, `R_s = 10 Omega + 15 Omega = 25 Omega`. Thus, `I_(min) = (12 V)/(25 Omega) = 0.48 A`. To obtain maximum current, the two resistors should be connected in parallel so that the equivalent resistance, `R_p = (15 Omega xx 10 Omega)/(15 Omega + 10 Omega) = 6 Omega`. Thus, `I_(max) = (12 V)/(6 Omega) = 2 A`.