InterviewSolution
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InterviewSolution
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Show how you would connect three resistors, each of resistance 6 Omega so that the combination has a resistance of (i) 9 Omega (ii) 4 Omega |
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Answer» Solution :It is given here that `R_1=R_2=R_3=6 OMEGA` To get a net resistance of `9 Omega` we join three resistances as shown in fig. 12.5 (a) Here resistances of PARALLEL COMBINATION of two `6 Omega` resistors is `1/R.=1/6+1/6=2/6=1/3implies R.=3 Omega` `therefore Net resistance=R.+6=3+6=9 Omega`  (ii) To get a net resistance of `4 Omega` we join three resistances as shown in fig.12.5 (b) Here the combination of two RESISTORES in series, having a combined resistances `R_0=6+6=12 Omega`, is joined in parallel with the THIRD resistances of `6 Omega` Hence, the net resistance will be `1/R=1/12+1/6=(1+2)/12=3/12=1/4 implies R= 4 Omega` |
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