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If, in Fig. \(34-2, R=20 \Omega, L=0.30 \mathrm{H}\), and \(\mathscr{E}=90 \mathrm{~V}\), what will be the current in the circuit \(0.050 \mathrm{~s}\) after the switch is closed? We are going to use the exponential equation for \(i\) given on p. 374 . The time constant for this circuit is \(L / R=0.015 \mathrm{~s}\), and \(i_{\infty}=\mathscr{/} R=4.5 \mathrm{~A}\). Then $$ i=i_{\infty}\left(1-e^{-t / L / R}\right)=(4.5 \mathrm{~A})\left(1-e^{-3.33}\right)=(4.5 \mathrm{~A})(1-0.0357)=4.3 \mathrm{~A} $$

Short Answer

Expert verified
The current in the circuit at 0.050 s is approximately 4.34 A.

Step by step solution

01

Identify Known Values

From the problem statement, we are provided with:1. Resistance \( R = 20 \, \Omega \)2. Inductance \( L = 0.30 \, \text{H} \)3. Voltage \( \mathscr{E} = 90 \, \text{V} \)4. Time \( t = 0.050 \, \text{s} \)Additional known values include:1. Time constant \( \tau = \frac{L}{R} = 0.015 \, \text{s} \)2. Final current \( i_{\infty} = \frac{\mathscr{E}}{R} = 4.5 \, \text{A} \)
02

Set Up the Exponential Equation

The equation for the current \( i(t) \) in an L-R circuit is given by:\[ i(t) = i_{\infty} (1 - e^{-t / \tau}) \]where \( \tau = \frac{L}{R} \) is the time constant and \( i_{\infty} \) is the final steady-state current.
03

Calculate Exponential Decay Term

Substitute the values \( t = 0.050 \, \text{s} \) and \( \tau = 0.015 \, \text{s} \) into the exponential decay part:\[ e^{-t / \tau} = e^{-0.050 / 0.015} = e^{-3.33} \]The approximate value of \( e^{-3.33} \) is \( 0.0357 \).
04

Substitute into the Current Equation

Substitute \( i_{\infty} = 4.5 \, \text{A} \) and \( e^{-3.33} = 0.0357 \) into the current equation:\[ i(t) = 4.5 \, \text{A} \times (1 - 0.0357) \]
05

Solve for Current at Specific Time

Calculate the current:\[ i(t) = 4.5 \, \text{A} \times 0.9643 = 4.34 \, \text{A} \]

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Key Concepts

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

Exponential Decay in an LR Circuit
Exponential decay refers to the process where the current in an LR circuit decreases over time after reaching a transient state. In an LR (inductor-resistor) circuit, this decay occurs as the circuit responds to changes in voltage after the switch is closed. The behavior of this current decay is modeled by the equation:
\[ i(t) = i_{\infty} \left(1 - e^{-t / \tau}\right)\]Here, \(i_{\infty}\) is the steady-state current, and \(\tau\) is the time constant. The term \(e^{-t/\tau}\) represents the exponential decay part.
  • Exponential decay is characterized by a rapid drop in current initially, slowing over time.
  • It is a gradual process that's asymptotic, meaning the current approaches a value indefinitely without reaching it.
  • This concept is crucial for understanding how inductors resist changes in current flow, affecting the overall energy dynamics inside the circuit.
Understanding exponential decay is essential for circuit analysis as it helps predict how quickly a circuit component like an inductor or resistor stabilizes to a new condition.
Time Constant and Its Significance
The time constant, often denoted by \(\tau\), is a pivotal parameter in analyzing an LR circuit. It effectively dictates the rate at which current changes occur and how swiftly a circuit reaches its steady state after the switch is closed.
  • The time constant is calculated as the ratio of inductance \(L\) to resistance \(R\):\[ \tau = \frac{L}{R}\]
  • It serves as a measure of how fast or slow the exponential decay of current happens in the circuit.
  • In our example, the time constant was calculated as \(0.015\, \text{s}\).
If the time constant is small, the circuit reaches its steady-state rapidly, indicating a fast response. A larger time constant means a slower response time.Due to its importance:
  • A time constant of \(\tau\) seconds means that after this period, the current will have completed about 63.2% of its decay towards the final value.
  • In practical applications, after approximately 5 time constants, the transient effects are considered negligible and the circuit is said to have reached its steady state.
Understanding the time constant is crucial for predicting the behavior of electrical circuits in a variety of engineering and physics applications.
Steady-State Current in the LR Circuit
The steady-state current, denoted as \(i_{\infty}\), is the current that flows through the circuit once all transient behaviors have subsided, and the circuit is in full equilibrium.
In our LR circuit, the steady-state current can be determined by Ohm’s Law once the inductive effects are negligible:
  • The formula used is:\[i_{\infty} = \frac{\mathscr{E}}{R}\]
  • Where \(\mathscr{E}\) is the electromotive force (emf) or the battery voltage and \(R\) is the resistance.
For our problem:
  • The given values result in a steady-state current of \(4.5\, \text{A}\).
  • It is the maximum current that would flow through the circuit if left undisturbed indefinitely.
The concept of steady-state current is fundamental to appreciate circuit behavior over a longer time frame. It's especially useful for scenarios where analysis of the final, unchanging current is required, such as in power calculations and ensuring component ratings are appropriate for continuous operation.

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Most popular questions from this chapter

A 2000-loop solenoid is wound uniformly on a long rod with length \(d\) and cross-section \(A\). The relative permeability of the iron is \(k_{m}\). On top of this is wound a 50-loop coil which is used as a secondary. Find the mutual inductance of the system. The flux through the solenoid is $$ \Phi_{M}=B A=\left(k_{M} \mu_{0} n I_{p}\right) A=\left(k_{M} \mu_{0} n I_{p} A\right)\left(\frac{2000}{d}\right) $$ This same flux goes through the secondary. We have, then, $$ \left|\mathscr{E}_{s}\right|=N_{s}\left|\frac{\Delta \Phi_{M}}{\Delta t}\right| \quad \text { and } \quad\left|\varepsilon_{s}\right|=M\left|\frac{\Delta i_{p}}{\Delta t}\right| $$ from which $$ M=N_{s}\left|\frac{\Delta \Phi_{M}}{\Delta i_{p}}\right|=N_{s} \frac{\Phi_{M}-0}{I_{p}-0}=50 \frac{k_{M} \mu_{0} I_{p} A(2000 / d)}{I_{p}}=\frac{10 \times 10^{4} k_{M} \mu_{0} A}{d} $$

A charged capacitor is connected across a \(10-\mathrm{k} \Omega\) resistor and allowed to discharge. The potential difference across the capacitor drops to \(0.37\) of its original value after a time of \(7.0 \mathrm{~s}\). What is the capacitance of the capacitor?

The mutual inductance between the primary and secondary of a transformer is \(0.30 \mathrm{H}\). Compute the induced emf in the secondary when the primary current changes at the rate of \(4.0 \mathrm{~A} / \mathrm{s}\).

A coil of resistance \(15 \Omega\) and inductance \(0.60 \mathrm{H}\) is connected to a steady 120 - \(\mathrm{V}\) power source. At what rate will the current in the coil rise \((a)\) at the instant the coil is connected to the power source, and \((b)\) at the instant the current reaches 80 percent of its maximum value? The effective driving voltage in the circuit is the \(120 \mathrm{~V}\) power supply minus the induced back emf, \(L(\Delta i / \Delta t)\). This equals the p.d. in the resistance of the coil: $$ 120 \mathrm{~V}-L \frac{\Delta i}{\Delta t}=i R $$ [This same equation can be obtained by writing the loop equation for the circuit of Fig. \(34-2(a)\). In doing so. remember that the inductance acts as a back emf of value \(L \Delta i / \Delta t .]\) (a) At the first instant, \(i\) is essentially zero. Then $$ \frac{\Delta i}{\Delta t}=\frac{120 \mathrm{~V}}{L}=\frac{120 \mathrm{~V}}{0.60 \mathrm{H}}=0.20 \mathrm{~mA} / \mathrm{s} $$ (b) The current reaches a maximum value of \((120 \mathrm{~V}) / R\) when the current finally stops changing (i.e., when \(\Delta i / \Delta t=0\) ). We are interested in the case when $$ i=(0.80)\left(\frac{120 \mathrm{~V}}{R}\right) $$ Substitution of this value for \(i\) in the loop equation gives $$ 120 \mathrm{~V}-L \frac{\Delta i}{\Delta t}=(0.80)\left(\frac{120 \mathrm{~V}}{R}\right) R $$ from which \(\quad \frac{\Delta i}{\Delta t}=\frac{(0.20)(120 \mathrm{~V})}{L}=\frac{(0.20)(120 \mathrm{~V})}{0.60 \mathrm{H}}=40 \mathrm{~A} / \mathrm{s}\)

Two neighboring coils, \(A\) and \(B\), have 300 and 600 turns, respectively. A current of \(1.5 \mathrm{~A}\) in \(A\) causes \(1.2 \times 10^{-4} \mathrm{~Wb}\) to pass through \(A\) and \(0.90 \times 10^{-4} \mathrm{~Wb}\) to pass through \(B\). Determine \((a)\) the self-inductance of \(A,(b)\) the mutual inductance of \(A\) and \(B\), and \((c)\) the average induced emf in \(B\) when the current in \(A\) is interrupted in \(0.20 \mathrm{~s}\).

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