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When two resistances connected in series and parallel their equivalent resistances are 15Omega and (56)/(15)Omega respectively. Find the individual resistances. |
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Answer» <P> Solution :`R_(s)=R_(1)+R_(2)=15Omega "" …(1)``R_(p)=(R_(1)R_(2))/(R_(1)+R_(2))=(56)/(15)Omega""…(2)` From equation (1) substituting for `R_(1)+R_(2)` in equation (2) `(R_(1)R_(2))/(15)=(56)/(15)Omega "" therefore R_(1)R_(2)=56` `R_(2)=(56)/(15)Omega ""...(3)` Substituting for `R_(2)` in equation (1) from equation (3) `R_(1)+(56)/(15)=15`then, `(R_(1)^(2)+56)/(R_(1))=15` `R_(1)^(2)+56=15R_(1) RARR R_(1)^(2)-15R_(1)+56=0` The above equation can be SOLVED USING factorisation. `R_(1)^(2)-8R_(1)-7R_(1)+56=0` `R_(1)(R_(1)-8)-7(R_(1)-8)=0 rArr (R_(1)-8) (R_(1)-7)=0` If `(R_(1)=8Omega)`, using in equation (1) `8+R_(2)=15 rArr R_(2)=15-8=7Omega` `R_(2)=7Omega` i.e, (when `R_(1)=8Omega, R_(2)=7Omega`) If `(R_(1)=7Omega)`, substituting in equation (1) `7+R_(2)=15 rArr R_(2)=8Omega`, i.e, (when `R_(1)=8Omega, R_(2)=7Omega`) |
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