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Two cells epsi_1 and epsi_2 in the given circuit diagram have an emf of 5 V and 9 V and internal resistance of 0.3 Omega and 1.2Omega respectively. Calculate the value of current flowing through the resistance of 3 Omega . |
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Answer» Solution :Here net EMF of CIRCUIT `epsi = epsi_2 - epsi_1 = 9- 5 = 4 V` and total resistance of the circuit `R = (6 xx 3)/(6 + 3)+ 4.5 + 0.3 + 1.2 = 8Omega` ` therefore ` Main circuit current `I= epsi/R = (4V)/(8Omega) = 0.5A` If current flowing through `3OMEGA` resistance be `I_1` , then current flowing through `6Omega` resistance will be `(0.5 - I_1)` and hence `3I_1 = 6 xx (0.5 - I_1) rArr I_1 = 0.33 A` |
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