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An inductor with an inductance of \(2.50 \mathrm{II}\) and a resistance of \(8.00 \Omega\) is connected to the terminals of a battery with an emf of \(6.00 \mathrm{~V}\) and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is \(0.500 \mathrm{~A}\) : (c) the current \(0.250 \mathrm{~s}\) after the circuit is closed; (d) the final steady-state current.

Short Answer

Expert verified
The answers are: (a) 2.4 A/s, (b) 0.6 A/s, (c) 0.45 A, (d) 0.75 A.

Step by step solution

01

Finding the initial rate of current increase

We can use the formula \(dI/dt = ε/L\), where ε is the electromotive force and L is the inductance. This will give us the initial rate of change of current as \(dI/dt = 6.00V / 2.50H = 2.4 A/s\).
02

Finding the rate of current increase when current is 0.500 A

The rate of increase of current at a certain moment is given by the formula \(dI/dt = ε/(L * R) * (1 - i/I)\), where R is the resistance, i is the instantaneous current and I is the steady state current. The steady state current can be calculated using Ohm's Law \(I = ε/R\). Substituting the known values: \(I = 6.00V / 8.00Ω = 0.75 A\). Now substitute all values in the earlier formula: \(dI/dt = 6/(2.5 * 8) * (1 - 0.500/0.75) = 0.6 A/s\)
03

Finding the current 0.250 s after circuit is closed

The current at any time after the circuit is closed is given by the formula \(i = I(1 - e^{-t/τ})\), with τ = L/R. Substituting L=2.50H, R=8.00Ω gives τ = 0.3125s. Then, substituting I=0.75A, t=0.250s in the current formula will give \(i = 0.75(1 - e^{-0.25/0.3125}) = 0.45 A\).
04

Finding the final steady-state current

As the circuit reaches its steady state, the potential difference across the inductor becomes zero and the circuit behaves like a simple resistive circuit. Hence, the final current can be achieved by applying Ohm's law, i.e., \(I = ε/R\), Substituting ε=6.00V and R=8.00Ω yields \(I = 0.75A\). This is the final steady state current.

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Key Concepts

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

Electromagnetic Induction
Electromagnetic induction is a fascinating phenomenon discovered by Michael Faraday in the 1830s. It explains how changing magnetic fields can induce electric currents in a conductor. This is the principle behind transformers, inductors, and many other devices in electric circuits.
When a magnetic field around a conductor changes, it induces a voltage (known as electromotive force, or emf) in the conductor. This effect ties directly to Lenz's Law, which states that the induced emf will always work to oppose the change that produced it. In simpler terms, the direction of the induced current is such that it opposes the change in magnetic field that generates it.
Inductors, like the one mentioned in the exercise, are components specifically designed to utilize electromagnetic induction. They can store energy temporarily in a magnetic field as current flows through them. This stored energy affects how quickly current can increase or decrease, as seen in the rate of current change calculations in the exercise. By using the formula \(dI/dt = \frac{ε}{L}\), we can determine how the current changes over time in response to an applied voltage.
Electric Circuits
Electric circuits are the paths through which electricity flows. They comprise various components including resistors, capacitors, inductors, batteries, and switches. Each component plays a specific role, creating a system that allows electrical energy to perform work, like lighting a bulb or running a motor.
In the exercise, we have a simple electric circuit involving an inductor, a resistor, and a battery. The battery provides the emf necessary to drive current through the circuit. The presence of the inductor introduces a delay in reaching steady-state current as it resists changes to the current using the principle of electromagnetic induction.
The design of an electric circuit often requires understanding the role of each component and how it interacts with others. Resistors limit the flow of current and dissipate energy as heat, while inductors can oppose changes in current, adding complexity to the circuit dynamics. Calculations involving the rate of current increase, as mentioned in the solved steps, are essential in predicting how the circuit will behave over time.
Ohm's Law
Ohm's Law is one of the fundamental principles in electric circuits. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. The relationship is given by the formula:
\[I = \frac{V}{R}\]
Where \I\ is the current in amperes (A), \V\ is the voltage in volts (V), and \R\ is the resistance in ohms (Ω). This law is integral to understanding how circuits function and is used extensively in calculations involving resistive elements.
In the exercise, Ohm's Law was used to determine the steady-state current when the inductor no longer affects the circuit. By substituting the battery's emf and the circuit's total resistance into the law, the exercise finds the steady-state current, which is key to understanding long-term circuit behavior. It's crucial to remember that while Ohm's Law is extremely useful, it applies primarily to linear, or ohmic, materials, where the resistance remains constant with varying current and voltage.

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Most popular questions from this chapter

(a) What would have to be the self-inductance of a solenoid for it to store \(10.0 \mathrm{~J}\) of energy when a \(2.00 \mathrm{~A}\) current runs through it? (b) If this solenoid's cross-sectional diameter is \(4.00 \mathrm{~cm}\), and if you could wrap its coils to a density of 10 coils \(/ \mathrm{mm}\), how long would the solenoid be? (See Exercise \(30.11 .)\) Is this a realistic length for ordinary laboratory use?

A \(7.50 \mathrm{nF}\) capacitor is charged to \(12.0 \mathrm{~V}\), then disconnected from the power supply and connected in scrics through a coil. The period of oscillation of the circuit is then measured to be \(8.60 \times 10^{-3} \mathrm{~s}\) Calculate: (a) the inductance of the coil: (b) the maximum charge on the capacitor; (c) the total cnergy of the circuit; (d) the maximum current in the circuit.

A resistor with \(R=30.0 \mathrm{~S}\) and an inductor with \(L=0.600 \mathrm{H}\) are connected in series to a hattery that has emf \(50.0 \mathrm{~V}\) and negligible internal resistance. At time \(t\) after the circuit is completed, the energy stored in the inductor is \(0.400 \mathrm{~J}\). At this instant, what is the voltage across the inductor?

CP CALC A cylindrical solcnoid with radius \(1.00 \mathrm{~cm}\) and length \(10.0 \mathrm{~cm}\) consists of 300 windings of AWG 20 copper wirc, which has a resistance per length of \(0.0333 \Omega / \mathrm{m}\). This solenoid is connected in series with a \(10.0 \mu \mathrm{F}\) capacitor, which is initially uncharged. A magnetic ficld dirccted along the axis of the solcnoid with strength \(0.100 \mathrm{~T}\) is switched on abruptly. (a) The solenoid may be considered an inductor and a resistor in series. Use Faraday's law to determine the average emf across the solenoid during the brief switch-on interval, and determine the nct charge initially deposited on the capacitor. (Sec Excrcisc \(29.4 .)\) (b) At time \(t=0\) the capacitor is fully charged and there is no current. How much time does it take for the capacitor to fully discharge the first time? (c) What is the frequency with which the current oscillates? (d) IIow much energy is stored in the capacitor at \(t=0 ?\) (e) How long does it take for the total cnergy stored in the circuit to drop to \(10 \%\) of that value?

A \(35.0 \mathrm{~V}\) battery with negligible intemal resistance, a \(50.0 \Omega\) resistor, and a \(1.25 \mathrm{mH}\) inductor with negligible resistance are all connected in series with an open switch. The switch is suddenly closed. (a) How long after closing the switch will the current through the inductor reach onehalf of its maximum value? (b) How long after closing the switch will the energy stored in the inductor reach one-half of its maximum value?

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