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The value of equivalent resistance for the circuit shown below is, A. `2.123 Omega`B. `5.123 Omega`C. `4.23 Omega`D. `6.283 Omega` |
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Answer» Correct Answer - B Step I Find out the resistance between the nodes P and Q. `R_(eq_(1))=R_(1)+(1)/(((1)/(R_(2))+(1)/(R_(3))))` or `R_(eq_(1))=1+(1)/(((1)/(2)+(1)/(3)))` or `R_(eq_(1))=1+(2xx3)/(5)` or `R_(eq_(1))=2.2 Omega` Step II Now, find the resistance between nodes Q and T. `R_(eq_(2))=R_(4)+(1)/(((1)/(R_(5))+(1)/(R_(6))+(1)/(R_(7))))+R_(8)` `= 1+(1)/(((1)/(2)+(1)/(3)+(1)/(4)))+1` `= 1+0.923+1` `rArr R_(eq_(2))=2.923 Omega` Step III As from circuit diagram, the equivalent resistance must be the series combination of `R_(eq_(1))` and `R_(eq_(2))`. `therefore " " R_(eq)=R_(eq_(1))+R_(eq_(2))` `= 2.2+2.923` `R_(eq)=5.123 Omega` |
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