The current flowing through an inductor of se | vedclass

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(D) The induced e.m.f. $(e)$ in an inductor of self-inductance $L$ is given by the formula: $e = -L \frac{dI}{dt}$.Here,the negative sign indicates Le

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$D$

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(D) The induced e.m.f. $(e)$ in an inductor of self-inductance $L$ is given by the formula: $e = -L \frac{dI}{dt}$.Here,the negative sign indicates Lenz's Law,which states that the induced e.m.f. opposes the change in current.If we consider the magnitude of the induced e.m.f. $|e| = L \frac{dI}{dt}$,we can see that $|e|$ is directly proportional to $\frac{dI}{dt}$.Since $L$ is a constant,the graph of $|e|$ versus $\frac{dI}{dt}$ will be a straight line passing through the origin with a slope equal to $L$.Looking at the given options,graph $(D)$ represents a linear relationship between $e$ and $dI/dt$ passing through the origin.Therefore,the correct graph is $(D)$.

The current flowing through an inductor of self-inductance $L$ is continuously increasing at a constant rate. The variation of induced e.m.f. $(e)$ versus $dI/dt$ is shown graphically by which graph?

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