Two inductors L1 (inductance 1 mH, internal resistance 3 ohm) and L2 (inductance 2 mH, internal resi

Q: Why is the current minimum at $t=0$?

At $t=0$, the inductors strongly oppose the sudden change in current, acting as open circuits. This leaves only the $12\,\Omega$ resistor to conduct current, resulting in the highest overall resistance and thus the minimum current.

Q: Why do we ignore the inductance values ($1\text{ mH}$ and $2\text{ mH}$) in our calculations?

The inductance values determine *how fast* the current reaches its steady state (the time constant), but they do not affect the initial ($0\text{ A}$) or final steady-state current values, which depend only on the resistances.

Q: What happens to the internal resistances of the inductors at steady state?

In the steady state, the inductors act as perfect conducting wires, but their internal resistances remain in the circuit. They act just like normal resistors in parallel.

Q: Could the maximum current occur at some time between $t=0$ and $t=\infty$?

No. In a DC circuit with inductors and resistors, the current grows exponentially from its minimum initial value to its maximum steady-state value. It never overshoots the steady-state value.

Q: How do we know the three branches are in parallel?

The problem states they are 'all connected in parallel across a $5\text{ V}$ battery'. This means each branch experiences the same $5\text{ V}$ potential difference.

Question

Two inductors $L_1$ (inductance $1	ext{ mH}$, internal resistance $3\,\Omega$) and $L_2$ (inductance $2	ext{ mH}$, internal resistance $4\,\Omega$), and a resistor $R$ (resistance $12\,\Omega$) are all connected in parallel across a $5	ext{ V}$ battery. The circuit is switched on at time $t = 0$. The ratio of the maximum to the minimum current $(I_{\max} / I_{\min})$ drawn from the battery is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a freshly wired circuit. We have a $5	ext{ V}$ battery connected in parallel to three distinct branches.

The first two branches are a bit special—they contain inductors with their own internal resistances. Specifically, branch one has a $1	ext{ mH}$ inductor with a $3\,\Omega$ resistance, and branch two has a $2	ext{ mH}$ inductor with a $4\,\Omega$ resistance.

The third branch is straightforward; it contains only a standard $12\,\Omega$ resistor.

Our mission is to find the ratio of the maximum current to the minimum current drawn from the battery after the switch is closed at $t = 0$. To crack this, we need to understand the dual personality of inductors at two critical moments in time.

### The Initial Moment: Minimum Current

Let's freeze time exactly at $t = 0$, the moment the switch is closed.

**Inductors are fundamentally opposed to change.** Because the current was zero before the switch was closed, the inductors will fight tooth and nail to keep it at zero.

In this initial transient state, they act as **open circuits**. This means absolutely no current can flow through the first two branches.

The entire burden of conducting current falls solely on the third branch with the $12\,\Omega$ resistor. Because the circuit has its highest effective resistance at this moment, the current drawn from the battery is at its absolute minimum.

Using Ohm's law, we can easily calculate this minimum current:

$$I_{\min} = \frac{\varepsilon}{R} = \frac{5}{12}	ext{ A}$$

### The Steady State: Maximum Current

Now, let's fast forward to a long time after the switch has been closed ($t 	o \infty$).

The circuit has settled down, and the current is no longer trying to change. Since inductors only oppose *changing* currents, they completely drop their guard in this steady state.

They now act as **short circuits**, or simple connecting wires. However, their internal resistances ($3\,\Omega$ and $4\,\Omega$) are still very much present.

At this point, current flows freely through all three parallel branches. Because the current has multiple paths to take, the overall equivalent resistance of the circuit is at its lowest, meaning the current drawn from the battery is at its maximum.

Let's calculate the equivalent resistance ($R_{eq}$) of these three parallel branches:

$$\frac{1}{R_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{R}$$

Substituting our values:

$$\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12}$$

To add these fractions, we find the common denominator, which is $12$:

$$\frac{1}{R_{eq}} = \frac{4 + 3 + 1}{12} = \frac{8}{12} = \frac{2}{3}\,\Omega^{-1}$$

With the equivalent resistance found, we apply Ohm's law one more time to find the maximum current:

$$I_{\max} = \varepsilon 	imes \frac{1}{R_{eq}} = 5 	imes \frac{2}{3} = \frac{10}{3}	ext{ A}$$

### Final Calculation

We have successfully navigated the extremes of the circuit's behavior. We found $I_{\min} = \frac{5}{12}	ext{ A}$ and $I_{\max} = \frac{10}{3}	ext{ A}$.

The final step is simply to find their ratio:

$$\frac{I_{\max}}{I_{\min}} = \frac{10/3}{5/12}$$

Flipping the denominator and multiplying:

$$\frac{I_{\max}}{I_{\min}} = \frac{10}{3} 	imes \frac{12}{5} = 2 	imes 4 = 8$$

**The ratio of the maximum to the minimum current is exactly 8.**

Notice how the actual inductance values ($1	ext{ mH}$ and $2	ext{ mH}$) were completely irrelevant to our final answer! They only dictate *how fast* the current transitions from minimum to maximum, not the boundary values themselves.

Enter Numerical Value:

Visualized Solution

$Circuit Setup$

$Critical Moments$

$t = 0 (Initial State)$

$I_{m i n} Calculation$

$t \to \infty (Steady State)$

$R_{e q} Setup$

$R_{e q} Calculation$

$I_{m a x} Calculation$

$Final Ratio$

$The Way Forward$

The Sigma Insight: Self and Mutual Inductance

Analyzing the Setup

The Initial Moment

The Steady State

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t→∞ (Steady State)

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

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

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

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

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