Two different coils have self-inductances L1 = 8 mH and L2 = 2 mH. The current in one coil is increa

Q: Why is the induced voltage proportional to inductance?

Because the rate of change of current is given as constant and identical for both coils. According to $V = L \frac{di}{dt}$, if $\frac{di}{dt}$ is constant, $V$ is directly proportional to $L$.

Q: Why do we use $P = V i$ for an inductor?

The instantaneous power delivered to any electrical component is the product of the voltage across it and the current flowing through it. For an inductor, this power goes into building the magnetic field.

Q: Is the energy stored in an inductor kinetic or potential?

It is magnetic potential energy. It represents the work done by the power source against the induced EMF to establish the current and the associated magnetic field.

Q: What happens if the current increases at different rates?

If the rates were different, say $\frac{di_1}{dt} \neq \frac{di_2}{dt}$, then the voltage ratio would depend on both the inductance ratio and the rate of change of current ratio.

Q: Why is the energy ratio $1/4$ even though $L_1$ is larger?

Although $L_1$ is 4 times larger than $L_2$, the current $i_1$ is $1/4$ of $i_2$. Since energy depends on the square of the current, the smaller current in the first coil significantly reduces its stored energy, resulting in a $1/4$ ratio.

Question

Two different coils have self-inductances $L_1= 8	ext{ mH}$ and $L_2 = 2	ext{ mH}$. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coils is the same. At that time, the current, the induced voltage and the energy stored in the first coil are $i_1, V_1$ and $W_1$ respectively. Corresponding values for the second coil at the same instant are $i_2, V_2$ and $W_2$ respectively. Then

Options:

(A) $\frac{i_1}{i_2} = \frac{1}{4}$

(B) $\frac{i_1}{i_2} = 4$

(C) $\frac{W_1}{W_2} = \frac{1}{4}$

(D) $\frac{V_1}{V_2} = 4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine two different coils, each acting as an inductor. The first coil has a self-inductance of $L_1 = 8	ext{ mH}$, and the second has $L_2 = 2	ext{ mH}$. We are told that the current in both coils is increasing at the exact same constant rate. This is a crucial piece of information because the rate of change of current, $\frac{di}{dt}$, directly determines the induced voltage across an inductor.

According to Faraday's law of induction, the induced voltage $V$ is given by the formula:
$$V = L \frac{di}{dt}$$

Since $\frac{di}{dt}$ is identical for both coils, the induced voltage is directly proportional to the inductance $L$.

### The Voltage Ratio
Let's find the ratio of the induced voltages. Because $V \propto L$, we can write:
$$\frac{V_1}{V_2} = \frac{L_1}{L_2}$$

Substituting the given values:
$$\frac{V_1}{V_2} = \frac{8}{2} = 4$$

This tells us that the voltage across the first coil is four times the voltage across the second coil. This confirms that option (d) is correct.

### The Current Ratio
Next, we are given a fascinating condition: at a specific instant, the power supplied to both coils is exactly the same. The electrical power $P$ delivered to any component is the product of its voltage and current:
$$P = V i$$

Since $P_1 = P_2$, we can equate the power for both coils:
$$V_1 i_1 = V_2 i_2$$

We want to find the ratio of the currents, $\frac{i_1}{i_2}$. Rearranging the equation gives:
$$\frac{i_1}{i_2} = \frac{V_2}{V_1}$$

We already know that $\frac{V_1}{V_2} = 4$, which means $\frac{V_2}{V_1} = \frac{1}{4}$. Therefore:
$$\frac{i_1}{i_2} = \frac{1}{4}$$

This confirms that option (a) is also correct.

### The Energy Stored
Finally, let's determine the magnetic potential energy stored in each coil. The energy $W$ stored in an inductor is given by:
$$W = \frac{1}{2} L i^2$$

To find the ratio of the energies, we divide the expression for the first coil by the expression for the second coil:
$$\frac{W_1}{W_2} = \frac{\frac{1}{2} L_1 i_1^2}{\frac{1}{2} L_2 i_2^2}$$

This simplifies to:
$$\frac{W_1}{W_2} = \left(\frac{L_1}{L_2}ight) \left(\frac{i_1}{i_2}ight)^2$$

Now, we just plug in the ratios we found earlier. We know $\frac{L_1}{L_2} = 4$ and $\frac{i_1}{i_2} = \frac{1}{4}$:
$$\frac{W_1}{W_2} = (4) \left(\frac{1}{4}ight)^2$$

Squaring the current ratio gives $\frac{1}{16}$:
$$\frac{W_1}{W_2} = 4 	imes \frac{1}{16} = \frac{1}{4}$$

This confirms that option (c) is correct as well.

**Conclusion:**
By systematically applying the formulas for induced voltage, power, and stored energy, we have found that options (a), (c), and (d) are all correct statements for this system of coils.

Select Answer:

Visualized Solution

$Induced Voltage Ratio$

$Current Ratio$

$Energy Stored Ratio$

The Sigma Insight: Self and Mutual Inductance

Analyzing the Setup Imagine two different coils, each acting as an inductor

The Voltage Ratio Let's find the ratio of the induced voltages

The Current Ratio Next, we are given a fascinating condition: at a specific instant, the power supplied to both coils is exactly the same

The Energy Stored Finally, let's determine the magnetic potential energy stored in each coil

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