Question

11. a. Can you tell which of the inductors in FIGURE Q30.11 has the larger current through it? If so, which one? Explain.

b. Can you tell through which inductor the current is changing more rapidly? If so, which one? Explain.

c. If the current enters the inductor from the bottom, can you tell if the current is increasing, decreasing, or staying the same? If so, which? Explain.

FIGURE Q30.11

Short answer

Part (a )It is not possible to calculate the current, and hence it is not possible to say which current is greater.

Part (b) The current through the second inductor is changing more rapidly than the first inductor.

Part (c )The current flowing through the inductors are decreasing.

Step-by-step solution

Part (a)

The voltage across an inductor is given by

$V=L\frac{dI}{dt}$

The above equation we can write that

$\begin{array}{l}\frac{dI}{dt}=\frac{V}{L}\\ \Rightarrow \int dI=\int \frac{V}{L}dt\\ \Rightarrow I=\frac{V}{L}t+C\end{array}$

Where C is the constant of integration. Now, since the values of i and C are not known, it is not possible to calculate current $I$through the given inductors.

Conclusion:

It is not possible to calculate the current, and hence it is not possible to say which current is greater.

Part (b)

We have seen that

$\frac{dI}{dt}=\frac{V}{L}$

For the first inductor we have

$\frac{d{I}_1}{dt}=\frac{2\text{V}}{2\text{H}}=1\text{A}/\text{s}$

And for the second inductor we have

$\frac{d{I}_2}{dt}=\frac{4\text{V}}{1\text{H}}=4\text{A}/\text{s}$

Hence, the current through the second inductor is changing rapidly.

Conclusion:

The current through the second inductor is changing more rapidly than the first inductor.

Part (c)

Part (c)

As it is already seen, the rate of change of current through inductor is given by

$\frac{dI}{dt}=\frac{V}{L}$

Now if the current is increasing the rate of change of current will be positive, that is the voltage and current will be in same direction. And, current is decreasing, the rate of change of current will be negative, that is the voltage and direction of current will be opposite.

Figure shows that the current is going upwards, but the voltage is downward. That is current is flowing opposite to the voltage, hence, $\frac{dI}{dt}$is negative. Hence, current is decreasing.

The current flowing through the inductors are decreasing.