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A coil of inductance \(300 \mathrm{mH}\) and resistance \(2 \Omega\) is connected to a source of voltage \(2 \mathrm{~V} .\) The current reaches half of its steady state value in [Kerala CET 2008] (a) \(0.05 \mathrm{~s}\) (b) \(0.1 \mathrm{~s}\) (c) \(0.15 \mathrm{~s}\) (d) \(0.3 \mathrm{~s}\)

Short Answer

Expert verified
The correct answer is (b) 0.1 seconds.

Step by step solution

01

Recognize the Expression for Current Increase in an RL Circuit

When a coil (inductor) is used in a circuit with a resistor, the current increase can be described by the equation: \[I(t) = I_0 (1 - e^{-t/\tau})\]where \( I_0 \) is the steady state current, \( t \) is time, and \( \tau \) is the time constant of the circuit, defined as \( \tau = \frac{L}{R} \). Here, \( L \) is the inductance and \( R \) is the resistance.
02

Calculate the Time Constant

The time constant \( \tau \) for the given RL circuit is:\[\tau = \frac{L}{R} = \frac{300\, \mathrm{mH}}{2\, \Omega} = \frac{0.3\, \mathrm{H}}{2\, \Omega} = 0.15\, \mathrm{s}\]
03

Determine the Steady State Current

The steady state current \( I_0 \) can be determined using Ohm's Law:\[I_0 = \frac{V}{R} = \frac{2\, \mathrm{V}}{2\, \Omega} = 1\, \mathrm{A}\]
04

Set the Equation for Half Steady State Current

We are given that the current reaches half its steady state value. Thus:\[\frac{I_0}{2} = I_0 (1 - e^{-t/\tau})\]Solving for \( t \), after substituting the values, gives us:\[\frac{1}{2} = 1 - e^{-t/0.15}\]Rearranging yields:\[e^{-t/0.15} = \frac{1}{2}\]
05

Solve for Time \( t \)

Take the natural logarithm on both sides to solve for \( t \):\[-\frac{t}{0.15} = \ln\left(\frac{1}{2}\right)\]\[t = -0.15 \times \ln\left(\frac{1}{2}\right)\]Using \( \ln(0.5) \approx -0.693 \):\[t = 0.15 \times 0.693 = 0.10395\, \mathrm{s}\]This value rounds to \(0.1\, \mathrm{s}\).
06

Conclusion

Based on the calculation, the correct answer is (b) \(0.1\, \mathrm{s}\). Thus, the time it takes for the current to reach half its steady state value is \(0.1\, \mathrm{s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inductance
Inductance is a fundamental concept in electrical circuits, particularly when discussing RL circuits. Inductance, denoted by the symbol \(L\), is measured in henrys (H) and represents a coil's ability to resist a change in the current flowing through it. This property arises because a changing current induces a voltage across the inductor, which opposes the change in current according to Lenz's Law.
In RL circuits, which consist of a resistor (R) and an inductor (L) connected in series to a voltage source, the inductor does not immediately allow the current to reach its maximum value. Instead, due to the inductance, the current gradually rises and eventually reaches a steady state. Inductance can be thought of as the electrical equivalent of inertia, a measure of how strongly the inductor resists changes to current flow.
  • In the example provided, the inductance is given as 300 mH, which means cases with high inductance see slower changes in current, because they can store more energy in their magnetic field.
Time Constant
In RL circuits, the time constant \(\tau\) is a key parameter that determines how quickly the circuit responds to changes in voltage. The time constant is defined as the ratio of the inductance \(L\) of the coil to the resistance \(R\) in the circuit: \(\tau = \frac{L}{R}\).
The time constant is expressed in seconds and provides a measure for how long it will take the current to rise to about 63.2% of its final steady state value after a voltage is applied. This is analogous to the concept of a half-life in radioactive decay, where a certain percentage of change consistently occurs over equal time intervals. The larger the time constant, the slower the reaction of the current will be.
  • Thus in our example, the time constant for the circuit is calculated as 0.15 seconds. After about one time constant, the current reaches approximately 63.2% of its maximum, demonstrating the effect of inductance on the response time of the circuit.
Steady State Current
Steady state current in RL circuits refers to the condition where the current remains constant over time after an initial transitory phase. In a circuit that has reached steady state, the inductor behaves like a short circuit with zero opposition to the flow of current.
  • To find the steady state current \(I_0\), Ohm's Law is applied: \(I_0 = \frac{V}{R}\), where \(V\) is the voltage source and \(R\) is the resistance. For example, when a 2 V source is connected to a 2 Ω resistor in the circuit, the steady state current will be 1 A.
In the context of RL circuits, reaching half of the steady state current involves solving the equation derived from the transient response of the inductor: \(I(t) = I_0 (1 - e^{-t/\tau})\). Understanding this concept is crucial because it tells us how the circuit transitions from one state to another, allowing engineers to predict circuit behavior under varying conditions.

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Most popular questions from this chapter

A cylindrical bar magnet is rotated about its axis. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then [NCERT Exemplar] (a) a direct current flows in the ammeter \(A\). (b) no current flows through the ammeter \(A\). (c) an altemating sinusoidal current flows through the ammeter \(A\) with a time period \(T=\frac{2 \pi}{\omega}\). (d) a time varying non-sinosoidal current flows through the ammeter \(A\).

The rails of a railway track insulated from each other and the ground are connected to a millivoltmeter. Find the reading of voltmeter, when a train travels with a speed of \(180 \mathrm{~km} / \mathrm{h}\) along the track. Given that the vertical component of earth magnetic field is \(0.2 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) and the rails are separated by \(1 \mathrm{~m}\) (a) \(10^{-4} \mathrm{~V}\) (b) \(10^{-2} \mathrm{~V}\) (c) \(10^{-3} \mathrm{~V}\) (d) \(1 \mathrm{~V}\)

Assertion In a series \(R-L-C\) circuit the voltage across resistor, inductor and capacitor are \(8 \mathrm{~V}, 16 \mathrm{~V}\) and \(10 \mathrm{~V}\) respectively. The resultant emf the circuit is \(10 \mathrm{~V}\). Reason Resultant emf of the circuit is given by the relation \(E=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}\)

A jet plane is travelling towards west at a speed of \(1800 \mathrm{~km} / \mathrm{h}\). What is the voltage difference developed between the ends of the wing having a span of \(25 \mathrm{~m}\), if the earth's magnetic field at the location has a magnitude of \(5 \times 10^{-4} \mathrm{~T}\) and the dip angle is \(30^{\circ}\). [NCERT] (a) \(2.1 \mathrm{~V}\) (b) \(3.1 \mathrm{~V}\) (c) \(4.1 \mathrm{~V}\) (d) \(5.2 \mathrm{~V}\)

A square loop of wire of side \(5 \mathrm{~cm}\) is lying on a horizontal table. An electromagnet above and to one side of the loop is turned on, causing a uniform magnetic field downwards at an angle of \(60^{\circ}\) to the vertical as shown in figure. The magnetic induction is \(0.50 \mathrm{~T}\). The average induced emf in the loop, if the field increases from zero to its final value in \(0.2 \mathrm{~s}\) is (a) \(5.4 \times 10^{-3} \mathrm{~V}\) (b) \(312 \times 10^{-3} \mathrm{~V}\) (c) 0 (d) \(0.25 \times 10^{-3} \mathrm{~V}\)

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