A 20 H inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at w

Q: Why does the current not reach its maximum value instantly?

An inductor opposes any change in current through it by inducing a back EMF. This causes the current to grow exponentially over time rather than jumping to its maximum value immediately.

Q: What is the difference between energy dissipated and the rate of energy dissipation?

Energy dissipated is the total heat produced over a time interval, while the rate of energy dissipation is the power at a specific instant, given by $i^2 R$.

Q: Why do we differentiate the magnetic energy to find its rate?

The magnetic energy stored is $U_L = \frac{1}{2}Li^2$. The rate at which this energy is being stored is how fast it changes with time, which is exactly what the time derivative $\frac{dU_L}{dt}$ represents.

Q: Can we cancel the current $i$ from both sides of the equation?

Yes, because we are looking for a time $t > 0$ when the current is actively growing and is not zero. Thus, dividing by $i$ is mathematically valid.

Q: What happens to the rates after a very long time?

As $t \to \infty$, the current reaches a steady maximum value, so $\frac{di}{dt} \to 0$. The rate of energy storage becomes zero, while the rate of heat dissipation reaches its maximum constant value.

Question

A $20	ext{ H}$ inductor coil is connected to a $10	ext{ ohm}$ resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is

Options:

(A) $\frac{2}{\ln 2}$

(B) $\frac{1}{2} \ln 2$

(C) $2 \ln 2$

(D) $\ln 2$

Sigma Percentile · Accepted Answer

The behavior of an L-R circuit is a fascinating interplay between the resistor, which dissipates energy, and the inductor, which stores it. Let's dive into the mathematics of this energy race!

### Analyzing the Setup

Imagine you have a battery, a resistor, and an inductor all connected in series. When you close the switch, the current doesn't just instantly jump to its maximum value.

Instead, the inductor fights back! It induces a back EMF that opposes the rise in current. Because of this, the current $i$ grows exponentially over time, governed by the classic equation:
$$i = \frac{E}{R}\left(1 - e^{-\frac{R}{L}t}ight)$$

As this current flows, two things happen simultaneously. First, the resistor starts heating up, dissipating energy at a rate given by Joule's law:
$$P_R = i^2 R$$

Second, the inductor starts building up its magnetic field, storing energy. The total energy stored is $U_L = \frac{1}{2}Li^2$. To find the rate at which this energy is being stored, we take the time derivative:
$$P_L = \frac{dU_L}{dt} = Li \frac{di}{dt}$$

### The Master Equation

The problem asks for the exact moment when these two rates are perfectly balanced. So, we set the rate of heat dissipation equal to the rate of magnetic energy storage:
$$i^2 R = Li \frac{di}{dt}$$

Since we are looking for a time $t > 0$ where the current $i$ is not zero, we can safely divide both sides by $i$. This simplifies our equation beautifully:
$$i R = L \frac{di}{dt}$$

Now, we need to substitute our expressions for the current and its derivative. Differentiating our current equation with respect to time gives:
$$\frac{di}{dt} = \frac{E}{L} e^{-\frac{R}{L}t}$$

Plugging this back into our simplified balance equation, we get:
$$\frac{E}{R}\left(1 - e^{-\frac{R}{L}t}ight) R = L \left(\frac{E}{L} e^{-\frac{R}{L}t}ight)$$

### Final Calculation

Notice how elegantly the terms cancel out! The resistance $R$ and the inductance $L$ disappear from the coefficients, leaving us with:
$$E\left(1 - e^{-\frac{R}{L}t}ight) = E e^{-\frac{R}{L}t}$$

We can also divide out the EMF $E$:
$$1 - e^{-\frac{R}{L}t} = e^{-\frac{R}{L}t}$$

Rearranging the terms, we find:
$$2e^{-\frac{R}{L}t} = 1 \implies e^{\frac{R}{L}t} = 2$$

Taking the natural logarithm of both sides unlocks the time $t$:
$$\frac{R}{L}t = \ln 2 \implies t = \frac{L}{R}\ln 2$$

Finally, we substitute the given values: $L = 20\,	ext{H}$ and $R = 10\,\Omega$.
$$t = \frac{20}{10}\ln 2 = 2\ln 2\,	ext{s}$$

**The required time is $2\ln 2$ seconds.**

Select Answer:

Visualized Solution

Current Growth $i (t)$

Power Dissipated $P_{R}$

Power Stored $P_{L}$

Equating $P_{R} = P_{L}$

Substituting $i$ and $\frac{d i}{d t}$

Solving for $t$

Final Value of $t$

The Sigma Insight: Self and Mutual Inductance

Analyzing the Setup

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

Current Growth .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i(t)

Power Dissipated .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PR​

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