Two identical inductors are connected in two different configurations $P$ and $Q$, where a time varying current $I(t)$ is flowing, as shown in the figure. The induced emf between points $a$ and $b$ for configuration $P$ is $E_P$ and that for configuration $Q$ is $E_Q$. The ratio $E_P/E_Q$ is:

[Neglect the effect of mutual inductance.]

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786186911/questions/y0vpdcdujlwz3y7ojuzh.png)

Question

Two identical inductors are connected in two different configurations $P$ and $Q$ , where a time varying current $I(t)$ is flowing, as shown in the figure. The induced emf between points $a$ and $b$ for configuration $P$ is $E_P$ and that for configuration $Q$ is $E_Q$ . The ratio $E_P/E_Q$ is:

[Neglect the effect of mutual inductance.]

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786186911/questions/y0vpdcdujlwz3y7ojuzh.png)

Accepted Answer

A. 1. For any inductor, the induced emf opposes the change in current and is given by:
$$\varepsilon = -L\frac{dI}{dt}$$

2. In configuration $P$, the two identical inductors combine to give an **effective inductance $L$** across $a$ and $b$. So:
$$E_P = -L\frac{dI(t)}{dt}$$

3. In configuration $Q$, the way the two inductors are connected gives an **effective inductance of $L/2$**. So:
$$E_Q = -\frac{L}{2}\frac{dI(t)}{dt}$$

4. Dividing the two expressions to compare them:
$$\frac{E_P}{E_Q} = \frac{L}{L/2} = 2$$

Hence, the answer is **option A**.