A uniformly wound solenoidal coil of self-inductance 1.8 x 10^-4 H and resistance 6 ohm is broken up

Q: Why does the inductance become half when the coil is cut in half?

The inductance of a solenoid is $L = \frac{\mu_0 N^2 A}{l}$. When cut in half, both the number of turns $N$ and the length $l$ are halved. Substituting $N/2$ and $l/2$ into the formula gives exactly half the original inductance.

Q: Why does the resistance become half?

Resistance is directly proportional to the length of the wire ($R = \rho l/A$). Cutting the coil in half means the wire length is halved, so the resistance is also halved.

Q: Can we calculate the time constant of each branch separately?

Yes! The time constant of each branch is $L'/R' = \frac{0.9 \times 10^{-4}}{3} = 3 \times 10^{-5}\text{ s}$. Since both branches are identical and connected in parallel, the overall time constant of the circuit remains the same.

Q: Why do we ignore the inductance when calculating steady state current?

In a DC circuit, the steady state is reached when the current becomes constant ($di/dt = 0$). Since the voltage across an inductor is $V_L = L(di/dt)$, it becomes zero, meaning the inductor acts as a perfect conductor (short circuit).

Q: What if the battery had internal resistance?

If the battery had internal resistance $r$, it would be in series with the equivalent resistance $R_{eq}$. The steady state current would then be $I_0 = \frac{V}{R_{eq} + r}$, and the time constant would be $\tau = \frac{L_{eq}}{R_{eq} + r}$.

Question

A uniformly wound solenoidal coil of self-inductance $1.8 	imes 10^{-4}	ext{ H}$ and resistance $6\ \Omega$ is broken up into two identical coils. These identical coils are then connected in parallel across a $15	ext{ V}$ battery of negligible resistance. The time constant for the current in the circuit is $\dots\dots	ext{ s}$ and the steady state current through the battery is $\dots	ext{ A}$.

Sigma Percentile · Accepted Answer

The problem asks us to find the time constant and the steady state current of a circuit formed by cutting a solenoid in half and connecting the two halves in parallel across a battery. Let's break this down step by step.

### Analyzing the Halved Solenoid

Imagine a long solenoid with an initial self-inductance $L = 1.8 	imes 10^{-4}	ext{ H}$ and a resistance $R = 6\ \Omega$. When we cut this solenoid exactly in half, we are essentially halving both its length $l$ and its total number of turns $N$.

Remember the formula for the inductance of a solenoid? It is given by:
$$L = \frac{\mu_0 N^2 A}{l}$$

Since the new number of turns is $N/2$ and the new length is $l/2$, the new inductance $L'$ becomes:
$$L' = \frac{\mu_0 (N/2)^2 A}{l/2} = \frac{1}{2} \frac{\mu_0 N^2 A}{l} = \frac{L}{2}$$

So, the new inductance is exactly half of the original, which gives $L' = 0.9 	imes 10^{-4}	ext{ H}$.

Similarly, the resistance of a wire is directly proportional to its length. Since the wire length is halved, the new resistance $R'$ is also halved:
$$R' = \frac{R}{2} = 3\ \Omega$$

### The Parallel Circuit Setup

Now, we take these two identical half-coils and connect them in parallel across a $15	ext{ V}$ battery.

To analyze this parallel combination, we need to find the equivalent inductance $L_{eq}$ and the equivalent resistance $R_{eq}$. Since both branches are identical, the equivalent values are simply half of the individual branch values.

For the equivalent inductance:
$$L_{eq} = \frac{L'}{2} = \frac{0.9 	imes 10^{-4}}{2} = 0.45 	imes 10^{-4}	ext{ H}$$

For the equivalent resistance:
$$R_{eq} = \frac{R'}{2} = \frac{3}{2} = 1.5\ \Omega$$

### Calculating the Time Constant

The time constant $	au$ of an L-R circuit dictates how quickly the current reaches its steady state. It is defined as the ratio of the equivalent inductance to the equivalent resistance.

Let's substitute our calculated values into the formula:
$$	au = \frac{L_{eq}}{R_{eq}} = \frac{0.45 	imes 10^{-4}}{1.5}$$

On solving this, we get:
$$	au = 0.3 	imes 10^{-4}	ext{ s} = 3 	imes 10^{-5}	ext{ s}$$

### The Steady State Current

Finally, let's determine the steady state current $I_0$ flowing through the battery.

In a DC circuit, once the steady state is reached (after a long time $t 	o \infty$), the current stops changing. Because the voltage across an inductor is proportional to the rate of change of current ($V_L = L \frac{di}{dt}$), the inductors behave as plain wires with zero resistance.

Therefore, the steady state current depends entirely on the battery voltage and the equivalent resistance of the circuit:
$$I_0 = \frac{V}{R_{eq}} = \frac{15}{1.5} = 10	ext{ A}$$

**Conclusion:**
We have successfully found the values for both blanks. The time constant is **$3 	imes 10^{-5}	ext{ s}$**, and the steady state current is **$10	ext{ A}$**.

Visualized Solution

$L and R of Halved Coil$

$Parallel Circuit Setup$

$L_{e q} and R_{e q}$

$Time Constant (τ)$

$Steady State Current (I_{0})$

$Conclusion$

The Sigma Insight: Self and Mutual Inductance

Analyzing the Halved Solenoid

The Parallel Circuit Setup

Calculating the Time Constant

The Steady State Current

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L and R of Halved Coil

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Steady State Current (I0​)

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