Question

In a certain circuit consisting of a constant e.m.f. E, an inductance L and a resistance R is closed, the current in it increases with time according to curve 1. After one parameter (E, L or R) is changed the increase in current follows curve 2, when the circuit is closed second time. Which parameter was changed and in what direction?

Answer 1

Hint
Use the slope of the given graph to know which one is more or less and then using the formula of self-inductance we can find its proportionality with the slope and hence we can know what changes have occurred in the parameters. Here,
 $ L\text{ }\alpha \text{ }\dfrac{1}{\dfrac{dI}{dt}} $
Self inductance is defined as the induction of a voltage in a current carrying wire when the current in the wire itself is changing. Self induction is the way in which single coils and chokes operate.

Complete step by step solution

Here, slope of the graph is $ \dfrac{dI}{dt} $
We can clearly see from the given graph that
Slope of curve 1 > slope of curve 2 i.e.
 $ {{\left( \dfrac{dI}{dt} \right)}_{1}}>{{\left( \dfrac{dI}{dt} \right)}_{2}} $
As we know,
 $ V=L\dfrac{dI}{dt} $
We can clearly see that L is inversely proportional to $ \dfrac{dI}{dt} $ i.e.
 $ L\text{ }\alpha \text{ }\dfrac{1}{\dfrac{dI}{dt}} $
Now if slope i.e. $ \dfrac{dI}{dt} $ is decreasing along the curve 2 that means that the self-inductance i.e. L is increasing here.
Or we can say
 $ {{L}_{2}}>{{L}_{1}} $
Hence, the changed parameter here is self-inductance L which has increased during the increase in current along the curve 2.

Note
Remember that the self-inductance is directly proportional to $ \dfrac{dt}{dI} $ which is the inverse of the slope in the given figure. Had there been the graph of time vs current with the same curves, the self-inductance would have been said to decrease with the decrease in slope of curve 2.