Question

Two resistors, with resistances 5 Ω and 10 Ω respectively are to be connected to a battery of emf 6 V so as to obtain: (i) minimum current flowing (ii) maximum current flowing.
(a) How will you connect the resistances in each case?
​(b) Calculate the strength of the total current in the circuit in two cases.

Answer 1

(a) In a parallel combination of resistances, the net resistance is less as compared to a series combination of resistances. As the current flow is indirectly proportional to the resistance of the circuit, so to obtain maximum current flowing in the circuit the given resistances should be connected in parallel combination and to obtain the minimum current flowing the given resistances should be connected in series combination.

(b) Case (i) For minimum current flow, the resistance of the circuit should be maximum. i.e. resistances should be connected in Series.

Net resistances of given resistances when connected in series = 5 Ω + 10 Ω = 15 Ω

The voltage of the battery = 6 V

Hence, current = $\frac{6\mathrm{V}}{15\mathrm{\Omega }}=0.4\mathrm{A}$

Case (ii) For maximum current flow, the resistance of the circuit should be minimum. i.e. resistances should be connected in parallel.

Net resistances of given resistances when connected in parallel = $\frac{5\mathrm{\Omega }\times 10\mathrm{\Omega }}{5\mathrm{\Omega }+10\mathrm{\Omega }}=\frac{10}{3}\mathrm{\Omega }$

The voltage of the battery = 6 V

Hence, current = $\frac{6\mathrm{V}}{{\displaystyle \frac{10}{3}}\mathrm{\Omega }}=1.8\mathrm{A}$