1.

What amount of heat will be generated in a coil of resistance `R` due to a charge q passing through it if the current in the coil a. decreases down to zero uniformly during a time interval `t_0`? b. decrases down to zero having its value every `t_0` seconds?

Answer» (a) As current `i` is linear function of time, and at `t = 0` and `Delta r`, it equals `i_(0)` and zero respectively, it may be represented as,
`i = i_(0) (1 - (t)/(Delta r))`
Thus `q = int_(0)^(Delta t) i dt = int_(0)^(Delta t) i_(0) (1 - (t)/(Delta r)) dt = (i_(0) Delta t)/(2)`
So, `i_(0) = (2q)/(Delta t)`
Hence, `i = (2q)/(Delta t) (1 - (t)/(Delta r))`
The heat generated.
`H = int_(0)^(Delta t) i^(2) R dt = int_(0)^(Delta t) [(2q)/(Delta r) (1 - (t)/(Delta r))]^(2) R dt = (4q^(2) R)/(3 Delta r)`
(b) Obviously the current through the coll is given by
`i = i_(0) ((1)/(2))^(t//Delta t)`
Then charge `q = int_(0)^(oo) idt = int_(0)^(oo) i_(0) 2^(-t//Delta t) dt = (i_(0) Delta t)/(In 2)`
So, `i_(0) = (q In 2)/(Delta t)`
And hence, heat generated in the circuit in the time interval `t[0, oo]`,
`H = int_(0)^(oo) i^(2) R dt = int_(0)^(oo) [(q In 2)/(Delta t)2^(-t//Deltat)]^(2) R dr = (-q In 2)/(2 Delta t) R`


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