Consider the following circuit `(Fig. 7.18)` where some resistances have been arranged in a definite order . With the given condition that heat produced by `6 Omega` resistance is `60 cals^(-1)` due to the current flowing throught it , find out the heat produced across `2 Omega` resistance in calorie per second.

Explanation

Same current flows through resistances connected in series . Heat produced in the ` 6 omega` resistance is `I^(2) R // J`. So
` 60 = (I^(2) xx 6)/( 4.2) or I^(2) or I = sqrt(42) A`
Now the voltage drop across `x and y` is
`( 6 + 12) sqrt( 42) = 18 sqrt( 42) V`
As this potential drop is same in every area of a parallel circuit , the potential drop across the upper part of the circuit ias the same . Therefore , current through the ` 2 Omega and 4 Omega` resistances is
`= (18 sqrt(42))//6 = 3 sqrt(42) A`
Hence , the heat produced across the ` 2 Omega` resistance is
`(I^(2)R)/( J) = ( 9 xx 42 xx 2) /( 4.2) = 180 cal`