In the figure below, the switches S1 and S2 are closed simultaneously at t = 0 and a current starts

Q: Why can we treat the two loops as completely independent?

The middle wire has zero resistance and zero inductance. Therefore, the potential difference between its top and bottom ends is always zero. This means the voltage from the left battery only drops across the left components, and the right battery only drops across the right components. They don't interfere with each other.

Q: How do we know the direction of $I_1$ and $I_2$?

Look at the battery symbols. The longer line represents the positive terminal. In the left loop, the positive terminal is up, driving a clockwise current ($I_1$). In the right loop, the positive terminal is down, which also drives a clockwise current ($I_2$).

Q: Why is the net current $I = I_1 - I_2$ and not $I_1 + I_2$?

Since both $I_1$ and $I_2$ are flowing clockwise in their respective loops, $I_1$ flows downwards through the shared middle wire, while $I_2$ flows upwards through it. Since the question defines $I$ as the downward current, we subtract the upward current ($I_2$) from the downward current ($I_1$).

Q: What happens if we don't ignore mutual inductance?

If mutual inductance were present, the changing magnetic field from the left inductor would induce an EMF in the right inductor, and vice versa. The loops would no longer be independent, and we would have to solve a system of coupled differential equations.

Q: How did $e^{-2\ln 2}$ become $1/4$?

Using the properties of logarithms, $-2\ln 2$ can be written as $\ln(2^{-2})$, which is $\ln(1/4)$. Since $e^{\ln x} = x$, $e^{\ln(1/4)}$ simplifies exactly to $1/4$.

Question

In the figure below, the switches $S_1$ and $S_2$ are closed simultaneously at $t = 0$ and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current $I$ in the middle wire reaches its maximum magnitude $I_{\max}$ at time $t = 	au$. Which of the following statements is (are) true ?

Options:

(A) $I_{\max} = \frac{V}{2R}$

(B) $I_{\max} = \frac{V}{4R}$

(C) $	au = \frac{L}{R} \ln 2$

(D) $	au = \frac{2L}{R} \ln 2$

Sigma Percentile · Accepted Answer

### The Illusion of Complexity

When you first look at this circuit, it seems like a tangled mess. Two loops, two batteries, two inductors, and a shared middle wire. Your brain might immediately jump to setting up complex Kirchhoff's Voltage Law (KVL) equations with multiple variables. But take a deep breath and look closer.

### Decoupling the Loops

Notice the middle wire. It has absolutely no components on it—no resistor, no inductor, no battery. It is an ideal, zero-resistance wire. Because of this, the potential difference across this middle wire is exactly zero.

What does this mean physically? It means the left loop and the right loop are completely independent of each other! They don't interact. The left battery only drives current through the left loop, and the right battery only drives current through the right loop. We have just turned one difficult problem into two very simple, standard $L-R$ circuit problems.

### The Battle of the Currents

Let's analyze the left loop. It's a simple series circuit with a battery $V$, a resistor $R$, and an inductor $L$. The battery's positive terminal is pointing upwards, which means it drives a clockwise current. Let's call this $I_1$. From our standard $L-R$ circuit theory, the growth of this current is given by:
$$I_1(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}ight)$$

Now, let's look at the right loop. It has a battery $V$, a resistor $R$, and an inductor $2L$. But pay close attention to the battery's polarity! The longer line is at the bottom, meaning the positive terminal is pointing downwards. This battery also drives a clockwise current, which we'll call $I_2$. The equation for this current is:
$$I_2(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{2L}t}ight)$$

Both of these currents pass through the shared middle wire. $I_1$ is flowing downwards, while $I_2$ is flowing upwards. The question asks for the net downward current $I$. Therefore, we must subtract the upward current from the downward current:
$$I(t) = I_1(t) - I_2(t)$$

Substituting our expressions, the constant $\frac{V}{R}$ terms cancel out beautifully, leaving us with:
$$I(t) = \frac{V}{R} \left( e^{-\frac{R}{2L}t} - e^{-\frac{R}{L}t} ight)$$

### Calculus to the Rescue

We need to find the maximum value of this current. In physics, whenever we need to maximize a function, we call upon calculus. We take the derivative of the current with respect to time and set it to zero:
$$\frac{dI}{dt} = 0$$

Differentiating our expression using the chain rule, we get:
$$\frac{V}{R} \left( -\frac{R}{2L} e^{-\frac{R}{2L}t} + \frac{R}{L} e^{-\frac{R}{L}t} ight) = 0$$

We can divide out the common factors and rearrange the terms:
$$\frac{R}{L} e^{-\frac{R}{L}t} = \frac{R}{2L} e^{-\frac{R}{2L}t}$$

Multiplying both sides by $\frac{L}{R}$ gives:
$$e^{-\frac{R}{L}t} = \frac{1}{2} e^{-\frac{R}{2L}t}$$

Or, more simply:
$$2 e^{-\frac{R}{L}t} = e^{-\frac{R}{2L}t}$$

To solve for $t$, we divide both sides by $e^{-\frac{R}{L}t}$:
$$2 = e^{\frac{R}{2L}t}$$

Taking the natural logarithm of both sides yields the time $	au$ at which the current is maximum:
$$\ln 2 = \frac{R}{2L} 	au \implies 	au = \frac{2L}{R} \ln 2$$
This confirms that option (d) is correct!

### The Final Calculation

Now that we have the exact time $	au$, we can find the maximum current $I_{\max}$ by plugging this time back into our current equation:
$$I_{\max} = \frac{V}{R} \left( e^{-\frac{R}{2L}\left(\frac{2L}{R} \ln 2ight)} - e^{-\frac{R}{L}\left(\frac{2L}{R} \ln 2ight)} ight)$$

Look at how perfectly the exponents simplify!
$$I_{\max} = \frac{V}{R} \left( e^{-\ln 2} - e^{-2\ln 2} ight)$$

Using the property of logarithms $e^{-x} = \frac{1}{e^x}$, we get:
$$I_{\max} = \frac{V}{R} \left( \frac{1}{2} - \frac{1}{4} ight)$$

$$I_{\max} = \frac{V}{4R}$$
This confirms that option (b) is also correct.

By simply recognizing that the loops were decoupled, we avoided a massive algebraic headache and solved the problem with elegance and precision!

Select Answer:

Visualized Solution

$Decoupling Loops$

$I_{1} (t) = \frac{V}{R} (1 - e^{- \frac{R}{L} t})$

$I_{2} (t) = \frac{V}{R} (1 - e^{- \frac{R}{2 L} t})$

$I_{n e t} = I_{1} - I_{2}$

$I (t) = \frac{V}{R} (e^{- \frac{R}{2 L} t} - e^{- \frac{R}{L} t})$

$\frac{d I}{d t} = 0$

$τ = \frac{2 L}{R} ln 2$

$I_{m a x} = I (τ)$

$I_{m a x} = \frac{V}{4 R}$

The Sigma Insight: Self and Mutual Inductance

The Illusion of Complexity

Decoupling the Loops

The Battle of the Currents

Calculus to the Rescue

The Final Calculation

Similar Questions

Consider the $L - R$ circuit shown in the figure. If the switch $S$ is closed at $t = 0$ , then the amount of charge that passes through the battery between $t = 0$ and $t = \frac{L}{R}$ is

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?

As shown in the figure, a battery of emf $ε$ is connected to an inductor $L$ and resistance $R$ in series. The switch is closed at $t = 0$ . The total charge that flows from the battery, between $t = 0$ and $t = t_{c}$ ( $t_{c}$ is the time constant of the circuit) is

In the circuit shown, The switch $S_{1}$ is closed at time $t = 0$ and the switch $S_{2}$ is kept open. At some later time ( $t_{0}$ ), the switch $S_{1}$ is opened and $S_{2}$ is closed. The behaviour of the current $I$ as a function of time ' $t$ ' is given by

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Decoupling Loops

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​(t)=RV​(1−e−LR​t)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​(t)=RV​(1−e−2LR​t)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Inet​=I1​−I2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I(t)=RV​(e−2LR​t−e−LR​t)

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τ=R2L​ln2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Imax​=I(τ)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Imax​=4RV​