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A network of resistor is connected to a 16 V battery with internal resistance of `1 Omega`, a shown in fig. Obtain the current in each resistor. |
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Answer» The total current I in the circuit is `I = (epsilon)/(R + r) = (16 v)/((7 + 1)Omega) = 2A` Consider the resistor between A and B. if `I_(1)` is the current in one of the `4 Omega` resistors and `I_(2)` the current in the other. `I_(1) xx 4 = I_(2) xx 4` That is `I_(1) = I_(2)`, which is otherwise obvious from the symmetry of the two arms. But `I_(1) + I_(2) = I = 2A`. Thus `I_(1) = I_(2) = IA` That is, current in each `4 Omega` resistor is 1A, Current in `1 Omega` resistor between B and C would be 2 A. Now, consider the resistances between C and D. If `I_(3)` is the current in the `12 Omega` resistor, and `I_(4)` in the `6 Omega` resistor, `I_(3) xx 12 = I_(4) xx 6`, i.e., `I_(4) = 2I_(3)` But, `I_(3) + I_(4) = I = 2A` Thus, `I_(3) = ((2)/(3)) A, I_(4) = ((4)/(3)) A` That is,the current in the `12 Omega` resistor is (2/3)A, while the current in the `6 Omega` resistor is (4/3) A. |
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