1.

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.

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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