A coil of inductance 2 H having negligible resistance is connected to a source of supply whose volta

Q: Why don't we use $V = IR$ here?

The problem explicitly states the coil has 'negligible resistance'. Therefore, the only opposition to the current comes from the inductor's self-induced emf, governed by $V = L \frac{dI}{dt}$.

Q: Can I just find the energy by integrating power $P = VI$?

Yes, absolutely! You can find $I(t)$ as a function of time, then integrate $P = V(t)I(t)$ from $t=0$ to $t=4$. It will give you the exact same result, though using $\frac{1}{2}LI^2$ is much faster once you know the final current.

Q: Why is the current initially zero?

An inductor opposes sudden changes in current. Since the circuit is connected at $t=0$ and the voltage starts at $0$ (since $V=3(0)=0$), the current must start from zero and build up gradually.

Q: What does 'energy stored in the coil' physically mean?

It refers to the energy stored in the magnetic field generated by the current flowing through the inductor's loops. It's analogous to kinetic energy in mechanics!

Q: What if the voltage was constant instead of $3t$?

If $V$ was constant, the current would increase linearly to infinity (since $R=0$). The $3t$ voltage makes the current increase quadratically instead!

Question

A coil of inductance $2	ext{ H}$ having negligible resistance is connected to a source of supply whose voltage is given by $V = 3t	ext{ V}$ (where, $t$ is in second). If the voltage is applied when $t = 0$, then the energy stored in the coil after $4	ext{ s}$ is ............ J.

Sigma Percentile · Accepted Answer

The journey to solving this problem begins with visualizing the physical setup. Imagine a simple electrical circuit. We have a coil, which acts as an inductor with a self-inductance of $L = 2	ext{ H}$. The problem explicitly states that this coil has negligible resistance, meaning we can treat it as an ideal inductor.

This inductor is connected to a voltage source, but there's a twist—this isn't your standard constant-voltage battery. The voltage is time-dependent, given by the equation $V = 3t$. This means the voltage starts at zero and ramps up linearly as time ticks forward. Our ultimate goal is to find the total energy stored in the magnetic field of this coil exactly $4	ext{ s}$ after the voltage is applied.

### The Master Equation

How does an ideal inductor respond to an applied voltage? According to **Faraday's Law of Induction**, an inductor opposes changes in current by inducing an electromotive force (emf). For an ideal inductor, the applied voltage $V$ is directly proportional to the rate of change of current $\frac{dI}{dt}$. The constant of proportionality is the inductance $L$.

This gives us our master equation:
$$V = L \frac{dI}{dt}$$

Let's substitute the values we know into this equation. We are given $V = 3t$ and $L = 2	ext{ H}$. 
$$3t = 2 \frac{dI}{dt}$$

### Integrating for Current

To find the energy stored, we first need to know the current $I$ flowing through the circuit at $t = 4	ext{ s}$. We can find this by separating the variables and integrating. Let's move $dt$ to the left side:
$$dI = \frac{3t}{2} dt$$

Now, we integrate both sides. Since the voltage is applied at $t = 0$, the initial current is zero. We want to find the current $I$ at $t = 4	ext{ s}$.
$$\int_0^I dI = \int_0^4 \frac{3t}{2} dt$$

The integral of $dI$ is simply $I$. On the right side, the integral of $t$ is $\frac{t^2}{2}$. 
$$I = \frac{3}{2} \left[ \frac{t^2}{2} ight]_0^4$$

Plugging in the upper limit of $4$:
$$I = \frac{3}{2} 	imes \frac{4^2}{2} = \frac{3}{2} 	imes \frac{16}{2} = \frac{3}{2} 	imes 8 = 12	ext{ A}$$

So, exactly $4	ext{ s}$ after the circuit is closed, a current of $12	ext{ A}$ is surging through the coil.

### Final Energy Calculation

Now for the grand finale. The energy $E$ stored in the magnetic field of an inductor is given by the classic formula:
$$E = \frac{1}{2} L I^2$$

We have all the pieces of the puzzle. We know $L = 2	ext{ H}$ and we just calculated $I = 12	ext{ A}$. Let's substitute these in:
$$E = \frac{1}{2} 	imes 2 	imes (12)^2$$

The $2$ in the numerator and denominator cancel out beautifully, leaving us with:
$$E = 12^2 = 144	ext{ J}$$

The total energy stored in the coil after $4	ext{ s}$ is **$144	ext{ J}$**. This problem is a beautiful demonstration of how time-varying voltages build up current and, consequently, magnetic energy in an inductive circuit!

Enter Numerical Value:

Visualized Solution

$V (t)$ \text{ and } $L$ \text{ Circuit}

\text{Faraday's Law of Induction}

\text{Substituting Values}

\text{Setting up the Integral}

\text{Calculating Current}

\text{Energy Stored in Inductor}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Self and Mutual Inductance

The Master Equation

Integrating for Current

Final Energy Calculation

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