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Which two resistors are connected in series with a cell of emf 2V and negligible internal resistance, a current of (2/5)A flows in the circuit. When the resistances are in parallel, the main current is (5/3)A. Calculate the resistances. |
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Answer» Solution :`I=E/(R+r)` or `R=V/I` Case (1): `R_S=2/(2//5)=5Omega` Case(2): `R_S=2/(5//3)=6/3 Omega` Simplification Arriving at `R_1 =3Omega` and `R_2=2Omega` Given , E=2V , r=0 , `I_s=2/5A` `I_P=5/3A R_1=? , R_2`=? (a)Resistors in SERIES :  `I_S=E/(R_S+r)` i.e., `2/5=2/((R_1+R_2)+0)` i.e., `R_1+R_2=5` ...(i) (B) Resistors in PARALLEL :  `I_p=E/(R_p+r)` i.e., `5/3=2/((R_1R_2)/(R_1+R_2) + 0)` i.e., `(R_1R_2)/(R_1+R_2)=6/5` `therefore R_1+R_2= 5 , R_1R_2=6` and `R_1+R_2=sqrt((5)^2 -4(6))=1`...(ii) Solving `R_1+R_2=5` `R_1-R_2=1` `2R_1=6` `R_1=3Omega` and `3-R_1=1` `R_2=2W` Hence `R_1=3Omega, R_2=2W` |
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