Q10 of 462 Page 22

In a simple circuit of resistance R, self inductance L and voltage E, the current i at any time t is given by If E is constant and initially no current passes through the circuit, prove that

We know that in a circuit of R, L and E we have,


⇒


We can see that it is a linear differential equation of the form


Where P = and Q =


I.F = e∫Pdt


= edt


=


Solution of the given equation is given by


i × I.F = ∫Q × I.F dt + c


⇒ i × = ∫ × dt + c


⇒ i × = ∫ × dt + c


⇒ i = + c ……(1)


Initially, there was no current


So, at i = 0, t = 0




Now, putting the value of c in equation (1)


i = –


i = (1 – )


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