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A circuit consists of three batteries of emf E_1 = 1 V, E_2= 2 V and E_3 = 3 V and internal resistances 1 Omega ,2Omega and 1 Omega respectively which are connected in parallel as shown in the figure. The potential difference between points P and Q is |
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Answer» Solution :Here ` E_1= 1 V, E_2=2 V ,E_3 =3V` ` r_1=1Omega, r_2= 2 Omega, r_3= 1 Omega` theeffectiveemfof thecircuitis ` E_("EFF")=((E_1)/( r_1)+(E_2)/(r_2) +(E_3)/(r_3))/((1)/(r_1) +(1)/(r_2) +(1)/(r_3) )= ((1)/(1) +(2)/(2) +3/1)/(1/1 +1/2 + 1/1 )= ( (10)/(2) )/(5/2) = 10/5 =2 V ` ` THEREFORE `potentialdifferencebetweenpointsP and Q` = E_("eff") = 2V ` |
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