The inductors of two LR circuits are placed next to each other, as shown in the figure. The values o

Q: Why do we ignore the energy dissipated as heat in the resistors?

The question specifically asks for the work done *against the induced EMF*, not the total work done by the battery. The total work done by the battery goes into both heating the resistor and building the magnetic field. The portion of work done specifically against the back-EMF of the inductor is exactly what gets stored as magnetic potential energy.

Q: How do we know the fluxes are aiding and not opposing?

By tracing the current direction from the positive terminal of each battery, we see both currents flow downwards through their respective coils. Assuming identical winding directions (which is the standard convention unless specified otherwise in the diagram), currents flowing in the same direction will produce fluxes that point in the same direction, thus aiding each other.

Q: What if the battery in the second circuit was reversed?

If the battery was reversed, the current $I_2$ would flow upwards through $L_2$. The fluxes would then oppose each other, and the mutual energy term would become negative, making the formula $U = \frac{1}{2}L_1 I_1^2 + \frac{1}{2}L_2 I_2^2 - M I_1 I_2$.

Q: Why is the steady state current independent of the inductance?

At steady state, the current is constant, meaning its rate of change $di/dt$ is zero. The voltage across an ideal inductor is $V = L(di/dt)$, which becomes zero. It acts like a simple wire with zero resistance, so only the actual resistors in the circuit limit the final current.

Q: Can the total magnetic energy ever be negative?

No, the total magnetic energy $\frac{1}{2}L_1 I_1^2 + \frac{1}{2}L_2 I_2^2 + M I_1 I_2$ is always non-negative. This is guaranteed by the physical constraint of coupling, where the mutual inductance $M$ can never exceed $\sqrt{L_1 L_2}$.

Question

The inductors of two LR circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistances, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, the total work done by the batteries against the induced EMF in the inductors by the time the currents reach their steady state values is________ mJ.

Sigma Percentile · Accepted Answer

### The Energy of Coupled Inductors

Imagine you are looking at two separate electrical circuits, each with its own battery, resistor, and inductor. At first glance, they seem completely independent. However, because the inductors are placed physically close to each other, they share a hidden connection: **mutual inductance**. This means the magnetic field created by one coil reaches over and influences the other.

The question asks us to find the total work done by the batteries against the induced EMFs by the time the currents reach their steady state. This might sound like a complex integration problem, but physics offers us a beautiful shortcut. The work done by the battery specifically against the induced EMF is not lost; it is entirely stored as **magnetic potential energy** in the inductors.

### Finding the Steady State Currents

Before we can calculate the energy, we need to know the final, steady state currents flowing through the circuits. What happens to an inductor after a long time in a DC circuit?

As the current stabilizes, its rate of change, $\frac{dI}{dt}$, becomes zero. Since the voltage across an ideal inductor is given by $V_L = L \frac{dI}{dt}$, the voltage drops to zero. The inductor effectively becomes a simple, resistance-less wire—a short circuit.

Therefore, the steady state current in each circuit is determined purely by Ohm's law, using only the battery voltage and the resistance:

For the first circuit:
$$I_1 = \frac{V_1}{R_1} = \frac{5	ext{ V}}{5\, \Omega} = 1	ext{ A}$$

For the second circuit:
$$I_2 = \frac{V_2}{R_2} = \frac{20	ext{ V}}{10\, \Omega} = 2	ext{ A}$$

### The Master Equation for Magnetic Energy

Now that we have the currents, we can calculate the stored magnetic energy. For a single isolated inductor, the energy is simply $\frac{1}{2}LI^2$. But for two coupled inductors, the total energy $U$ is the sum of their individual self-energies plus the energy arising from their mutual interaction:

$$U = \frac{1}{2}L_1 I_1^2 + \frac{1}{2}L_2 I_2^2 \pm M I_1 I_2$$

Here is the critical catch: **Should we use a plus or a minus sign for the mutual energy term?**

To decide, we must look at the direction of the magnetic fluxes. Let's trace the current from the positive terminal of each battery. In the left circuit, $I_1$ flows downwards through the coil $L_1$. In the right circuit, $I_2$ also flows downwards through the coil $L_2$. Assuming the coils are wound in the same standard direction, currents flowing in the same physical direction will produce magnetic fluxes that point in the same direction.

Because the fluxes **aid** each other, the mutual inductance increases the total stored energy. Therefore, we must use the **positive sign**.

### Final Calculation

Let's substitute our known values into the master equation. To keep things simple, we will leave the inductances in millihenries (mH), which will naturally give us our final energy in millijoules (mJ).

$$U = \frac{1}{2}(10)(1)^2 + \frac{1}{2}(20)(2)^2 + (5)(1)(2)$$

Breaking it down term by term:
- Self-energy of coil 1: $\frac{1}{2} 	imes 10 	imes 1 = 5	ext{ mJ}$
- Self-energy of coil 2: $\frac{1}{2} 	imes 20 	imes 4 = 40	ext{ mJ}$
- Mutual energy: $5 	imes 2 = 10	ext{ mJ}$

Adding them all together:
$$U = 5 + 40 + 10 = 55	ext{ mJ}$$

The total work done by the batteries against the induced EMFs is exactly **$55	ext{ mJ}$**.

Enter Numerical Value:

Visualized Solution

$W_{battery} = U_{magnetic}$

$t \to \infty ⟹ \frac{d I}{d t} = 0$

$I_{1} = \frac{V _{1}}{R _{1}}$

$I_{2} = \frac{V _{2}}{R _{2}}$

$U = \frac{1}{2} L_{1} I_{1}^{2} + \frac{1}{2} L_{2} I_{2}^{2} \pm M I_{1} I_{2}$

$U = \frac{1}{2} L_{1} I_{1}^{2} + \frac{1}{2} L_{2} I_{2}^{2} + M I_{1} I_{2}$

$U = \frac{1}{2} (10) (1)^{2} + \frac{1}{2} (20) (2)^{2} + (5) (1) (2)$

$U = 55 mJ$

The Sigma Insight: Self and Mutual Inductance

The Energy of Coupled Inductors

Finding the Steady State Currents

The Master Equation for Magnetic Energy

Final Calculation

Similar Questions

In the figure shown, a circuit contains two identical resistors with resistance $R = 5Ω$ and an inductance with $L = 2 mH$ . An ideal battery of $15 V$ is connected in the circuit. What will be the current through the battery long after the switch is closed?

The self-induced emf of a coil is 25 V. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is

Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Wbattery​=Umagnetic​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t→∞⟹dtdI​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​=R1​V1​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​=R2​V2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U=21​L1​I12​+21​L2​I22​±MI1​I2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U=21​L1​I12​+21​L2​I22​+MI1​I2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U=21​(10)(1)2+21​(20)(2)2+(5)(1)(2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U=55 mJ