The current flowing through the inductor of self inductance 'L' is continuously increasing at constant rate. The variation of induced e.m.f. (e) versus \(\frac{dI}{dt}\) is shown graphically by figure
Show Hint
Write e = -L dI/dt. It is a straight line through the origin with negative slope.
Step 1: Recall the law:
For an inductor with self inductance \(L\), the induced e.m.f. is \(e = -L\frac{dI}{dt}\). The minus sign is Lenz's law: the induced e.m.f. opposes the change in current.
Step 2: Compare with a straight line:
Put \(y = e\) and \(x = dI/dt\). Then \(e = (-L)\,\frac{dI}{dt}\) has the form \(y = mx\) with slope \(m = -L\). So the graph is a straight line through the origin, and the slope is negative because \(L\) is positive.
Step 3: Read the figures:
In the figure, graph A is a straight line through the origin with positive slope. Graph B is a horizontal line (e does not change with dI/dt). Graph C is a straight line through the origin that falls into the negative e region as dI/dt increases. Graph D is a curve.
Step 4: Match:
Only graph C is a straight line from the origin with negative e for positive dI/dt. Graph A has the wrong sign, graph B would mean e is independent of dI/dt, and graph D would mean e is not proportional to dI/dt.
Final Answer:
The e versus dI/dt graph is a straight line of slope \(-L\), which is figure C.
\[ \boxed{\text{C}} \]
