Electrical Engineering ⇒ Topic : Closing and Breaking an Inductive Circuit

Sachin
 
Closing and Breaking an Inductive Circuit Consider an inductive circuit shown in Fig. (a). When switch S is closed, the current increases gradually and takes some time to reach the final value. The reason the current does not build up *instantly to its final value is that as the current increases, the selfinduced e.m.f. in L opposes the change in current (Lenz's Law). Suppose at any instant, the current is i and is increasing at the rate of di/dt. Then, V = V_{R}+V_{L} _{} _{ figure (a)} _{} As the current increases, v_{R }(= iR) increases and V_{L}, decreases since V is constant. The decrease in v_{L}(= L di/dt) means that di/dt decreases because L is constant. The result is that after some time, di/dt becomes zero and so does the self induced e.m.f. v_{L}(= L di/dt). At this stage, the current attains the final fixed value I given by _{} Thus, when a d.c. circuit containing inductance is switched on, the current takes some time to reach the final value I (= V/R). Note that the role of inductance is to delay the change; it cannot prevent the current from attaining the final value. Similarly, when an inductive circuit is opened, the current does not jump to zero, but falls gradually. In either case, the delay in change depends upon the values of L and R as explained in the next article  
 
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