Problem 47 - Master Resource Book in JEE Main Physics

Question

A uniformly wound solenoidal coil of self-inductance \(1.8 \times 10^{-4} \mathrm{H}\) and resistance \(6 \Omega\) is broken up into two identical coils. These identical coils are then connected in parallel across a \(12 \mathrm{~V}\) battery of negligible resistance. The time constant of the current in the circuit and the steady state current through battery is (a) \(3 \times 10^{-5} \mathrm{~s}, 8 \mathrm{~A}\) (b) \(1.5 \times 10^{-5}\) s, \(8 \mathrm{~A}\) (c) \(0.75 \times 10^{-4} s, 4 \mathrm{~A}\) (d) \(6 \times 10^{-5} \mathrm{~s}, 2 \mathrm{~A}\)

Step-by-Step Solution

Answer

The time constant is \(3 \times 10^{-5} \mathrm{s}\) and the steady state current is \(8 \mathrm{~A}\). Choose (a).

Step 1: Determine the Inductance and Resistance of the Split Coils

When the solenoidal coil is split into two identical coils, each coil will have half the self-inductance and half the resistance of the original coil. Therefore, the inductance of each coil is \( \frac{1.8 \times 10^{-4}}{2} = 9 \times 10^{-5} \mathrm{H} \), and the resistance is \( \frac{6}{2} = 3 \Omega \).

Step 2: Calculate Equivalent Inductance and Resistance in Parallel

When two coils are connected in parallel, the equivalent inductance \( L_{eq} \) is given by \( \frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2} \). For the two identical coils \( L_1 = L_2 = 9 \times 10^{-5} \mathrm{H} \), the equivalent inductance is \[ L_{eq} = \frac{L}{2} = \frac{9 \times 10^{-5}}{2} = 4.5 \times 10^{-5} \mathrm{H}. \] The equivalent resistance for parallel resistances \( R_1 = R_2 = 3 \Omega \) is \[ R_{eq} = \frac{R}{2} = \frac{3}{2} = 1.5 \Omega. \]

Step 3: Determine the Time Constant of the Circuit

The time constant \( \tau \) of an LR circuit is given by the formula \( \tau = \frac{L}{R} \). Here, substitute \( L = 4.5 \times 10^{-5} \mathrm{H} \) and \( R = 1.5 \Omega \): \[ \tau = \frac{4.5 \times 10^{-5}}{1.5} = 3 \times 10^{-5} \mathrm{s}. \]

Step 4: Calculate the Steady State Current

In the steady state, the inductors behave like short circuits, so the total resistance of the circuit is \( R_{eq} = 1.5 \Omega \). Using Ohm's law, the steady state current \( I \) is given by \( I = \frac{V}{R} \), where \( V = 12 \mathrm{~V} \). Therefore: \[ I = \frac{12}{1.5} = 8 \mathrm{~A}. \]

Step 5: Identify the Correct Answer

From the calculation, the time constant is \( 3 \times 10^{-5} \mathrm{s} \) and the steady state current is \( 8 \mathrm{~A} \). Therefore, the correct choice is (a) \( 3 \times 10^{-5} \mathrm{s}, 8 \mathrm{~A} \).

Key Concepts

Self-Inductance

In an LR circuit, **self-inductance** is a critical property of the coil. It is a measure of a coil's ability to resist changes in current flowing through it. When a current flows through a coil, it generates a magnetic field around it. If the current changes, the magnetic field also changes, and by Faraday's law of electromagnetic induction, a voltage is induced that opposes the change in current. This property is known as self-inductance and is measured in henries (H).
The self-inductance of a uniformly wound solenoidal coil, when split into two identical coils, is halved for each coil. This is because the magnetic field characteristics change when a single coil is divided. For example, in our exercise, the original coil had a self-inductance of \( 1.8 \times 10^{-4} \mathrm{H} \). When split, the self-inductance of each coil becomes \( 9 \times 10^{-5} \mathrm{H} \).
Understanding self-inductance helps in comprehending how coils resist changes and store energy in their magnetic fields. This is fundamental to analyzing circuits involving inductors, especially in transient states, where currents change rapidly.

Time Constant Calculation

The **time constant** in an LR circuit is a vital factor that determines how quickly the circuit responds to changes in voltage or current. It is essentially the time it takes for the current to reach approximately 63.2% of its maximum value after the voltage is applied. Mathematically, the time constant \( \tau \) is defined as:
\[ \tau = \frac{L}{R} \]
where \( L \) is the inductance, and \( R \) is the resistance. This formula illustrates why both the inductance and resistance in a circuit are critical: more inductance or resistance means a slower response time.
In the given exercise, when the two identical coils are connected in parallel, the equivalent inductance and resistance also change. The equivalent inductance becomes \( 4.5 \times 10^{-5} \mathrm{H} \) and the resistance \( 1.5 \Omega \). Using these values, the time constant is calculated as \( 3 \times 10^{-5} \mathrm{s} \). This time constant describes how quickly the current will rise in this particular LR circuit when energized.

Steady State Current

In electrical circuits, achieving a **steady state** means the current and voltage become constant over time, usually after initial transients have settled. For an LR circuit, once equilibrium is reached, the inductive effects diminish, and the inductor behaves like a direct wire with zero impedance. As a result, the circuit's behavior depends solely on resistance.
To find the steady state current in our scenario, Ohm's law is applied:
\[ I = \frac{V}{R} \]
With a voltage of 12 V and an equivalent resistance of 1.5 \( \Omega \), the current calculates to \( 8 \mathrm{~A} \). This steady state current indicates how much current will continually flow through the circuit after it stabilizes.
Understanding steady state conditions is crucial for designing circuits that will operate consistently and predictably over time. It helps engineers determine the correct current ratings for components to avoid overheating and ensure efficient operation.