InterviewSolution
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InterviewSolution
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Two resistors, with resistance 5 Omega and 10 Omega resistively are to be connected to a battery of emf 6V so as to obtain: (i) Minimum current flowing (ii) maximum current flowing (a) How will you connect the resistance in each case? (b) Calculate the strength of the total current in the circuit in the two cases. |
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Answer» Solution :(a) (i) Here `R_1=5 Omega ,R_2=10 Omega and V=6V` (a) (i) For minimum current FLOWING in the circuit resistance `R_1 and R_2` should be connected in series. (ii) For maximum current flowing in the circuit resistances `R_1 and R_2` should be connected in PARALLEL. (B) (i) In series COMBINATION `R_s=R_1+R_2=5 Omega+10 Omega=15 Omega` `therefore` MInimum current `I_(min)=V/R_s=(6V)/(15 Omega)=0.4A` (ii) In parallel combination `1/R_p=1/5+1/10 implies R_p=10/3 Omega` `therefore` Maximum current `I_(max)=V/R_p=(6V)/(10/3 Omega)=1.8A` |
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