Write and explain Kirchhoff's second law (Loop rule).

Explanation

`rArr` The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.
OR
For any closed loop the algebraic sum of product of resistances and respective current flowing through them is equal to algebraic sum of emf applied along the loop.

`rArr` As shown in figure, ABCDEA is closed circuit. In different branches resistors are `R_(1), R_(2), R_(3) , R_(4) and R_(5)` and corresponding current are `I_(1) , I_(2), I_(3) , I_(4) and I_(5)` respectively.
`rArr` In the circuit cell `epsilon_(1) and epsilon_(2)` are connected.
`rArr` Potential at A, B, C, D and E be `V_(A) , V_(B) , V_(C) , V_(D) and V_(E)` respectively.
`rArr` If direction of motion current potential is considered as negative and in opposite directio, the potential is considered positive.
`rArr` When we move from negative to positive terminal of battery then emf is considered positive and from positive to negative termina emf is considered as negative.
`rArr` Starting from point A moving in clockwise direction and write potential of various point then it will be as follows :
`rArr` Potential at point A `V_(A) = V_(A)`
moving from A to B,
potential at B `V_(B) = V_(A) = I_(1) R_(I) + epsilon_(1)`
moving from B to C,
potentail at C, `V_(C) = V_(A) - I_(1) R_(1) + epsilon_(1) + I_(2) R_(2)`
moving from C to D,
`V_(D) = V_(A) - I_(1) R_(1) + epsilon_(1) + I_(2) R_(2) - epsilon_(2) - I_(3) R_(3)`
moving from D to E,
`V_(A) = V_(A) - I_(1) R_(I) + epsilon_(1) + I_(2) R_(2) - epsilon_(2) - I_(3) R_(3) + I_(4) R_(4) + I_(5) R_(5)`
`therefore I_(1) R_(I) - I_(2) R_(2) + I_(3) R_(3) - I_(4) R_(4) - I_(5) R_(5) = - epsilon_(1) + epsilon_(2)`
`therefore sum_(i=l)^(n) I_(i) R_(i) = sum_(i=l)^(n) epsilon_(I) `
`rArr ` This law is law of conservation of energy.