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Chapter 30: Problem 25

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?

Short Answer

Expert verified
The voltage across the inductor at this instant is 20.0 V

Step by step solution

01

Determine the current

First, we need to determine the current flowing through the inductor. Since we know the energy stored in the inductor is 0.400 J and we have the expression for the energy stored in an inductor, we can arrange it to solve for the current: \(I=\sqrt{\frac{2U}{L}}\) This gives us the current, \(I\), as \(I=\sqrt{\frac{2*0.400 \mathrm{~J}}{0.600 \mathrm{~H}}}=1.0 \mathrm{~A}\)
02

Calculate the voltage across the resistor

Next, we can calculate the voltage drop across the resistor using Ohm's Law (V=IR). Plugging in our values gives us \(V_R=I*R=1.0 \mathrm{~A} * 30.0 \mathrm{~\Omega}=30.0 \mathrm{~V}\)
03

Calculate the voltage across the inductor

We know that the battery voltage is equal to the sum of the voltage drops across the inductor and the resistor (since they are in series): \(V=V_R+V_L\). Therefore, we can use this to solve for the voltage across the inductor: \(V_L=V-V_R=50.0 \mathrm{~V} - 30.0 \mathrm{~V}=20.0 \mathrm{~V}\)

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Key Concepts

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

Resistor
A resistor is a fundamental component in electrical circuits. It limits the flow of electric current, helping to control the voltage and current in the circuit. The resistance is measured in ohms (惟), and a resistor's ability to restrict current flow is quantified by its resistance value. Resistance in a circuit is crucial as it directly impacts how energy is distributed across other components, like the inductor.
For example, in our original exercise, a resistor with a resistance of 30.0 惟 is used. The resistor absorbs a portion of the voltage from the battery, resulting in a specific voltage drop across its terminals.
Ohm's Law
Ohm鈥檚 Law is a key principle in electrical engineering and physics, often expressed as:
  • \[ V = IR \]
where \( V \) is voltage, \( I \) is current, and \( R \) is resistance. This simple formula allows you to calculate the relationship between voltage, current, and resistance in any part of a circuit.
Applying this law to the given exercise, we can calculate the voltage across the resistor by multiplying the current, 1.0 A, by the resistance, 30.0 惟, giving us a voltage of 30.0 V. This demonstrates how Ohm's Law provides a straightforward way to determine values in a circuit.
Series circuit
A series circuit is one where components are connected end-to-end, forming a single pathway for current flow. The same current flows through each component, and the total voltage around the circuit is the sum of the individual voltage drops across each component.
In the given exercise, the resistor and the inductor are in series with the battery. This setup ensures the current flowing through the resistor is the same as the current through the inductor. As such, the total voltage supplied by the battery is divided between the resistor and the inductor. This fundamental concept is why we can sum the voltage across the resistor and inductor to match the battery's emf.
Voltage across inductor
The voltage across an inductor in a circuit is determined by the changing current flowing through it. In our exercise, the battery's emf of 50 V is shared between the resistor and the inductor due to their series connection.
The voltage across the inductor can be calculated once the voltage drop across the resistor is known. Using the formula:
  • \[ V_L = V_{ ext{battery}} - V_R \]
we find the inductor's voltage as 20 V, after subtracting the resistor's voltage of 30 V from the battery's total voltage of 50 V. Understanding how to calculate this voltage is crucial for solving series circuit problems involving inductors.

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Most popular questions from this chapter

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?

\(\operatorname{An} L-C\) circuit containing an \(80.0 \mathrm{ml}\) inductor and a \(1.25 \mathrm{nF}\) capacitor oscillates with a maximum current of 0.750 A. Calculate: (a) the maximum charge on the capacitor and (b) the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time \(t=0,\) calculate the energy stored in the inductor after \(2.50 \mathrm{~ms}\) of oscillation.

An inductor used in a de power supply has an inductance of \(12.0 \mathrm{H}\) and a resistance of \(180 \Omega\). It carries a current of \(0.500 \mathrm{~A}\). (a) What is the energy stored in the magnetic field? (b) At what rate is thermal energy developed in the inductor? (c) Does your answer to part (b) mean that the magnetic-field energy is decreasing with time? Explain.

DATA To investigate the properties of a large industrial solenoid, you connect the solenoid and a resistor in series with a battery. Switches allow the battery to be replaced by a short circuit across the solenoid and resistor. Therefore Fig. 30.11 applies, with \(R=R_{\text {ext }}+R_{L}\). where \(R_{L}\) is the resistance of the solenoid and \(R_{\text {ext }}\) is the resistance of the series resistor. With switch \(S_{2}\) opcn, you close switch \(S_{1}\) and kecp it closed until the current \(i\) in the solenoid is constant (Fig. 30.11 ). Then you close \(S_{2}\) and open \(S_{1}\) simultaneously, using a rapid- response switching mechanism. With high-speed clectronics you measure the time \(t_{\text {balf }}\) that it takes for the current to decrease to half of its initial value. You repeat this measurement for several values of \(R_{\text {ext }}\) and obtain these results: \begin{tabular}{l|lllllll} \(R_{\text {ext }}(\Omega)\) & 3.0 & 4.0 & 5.0 & 6.0 & 7.0 & 8.0 & 10.0 & 12.0 \\\ \hline\(t_{\text {half }}(\mathrm{s})\) & 0.735 & 0.654 & 0.589 & 0.536 & 0.491 & 0.453 & 0.393 & 0.347 \end{tabular} (a) Graph your data in the form of \(1 / t_{\text {halt }}\) versus \(R_{e x l}\). Explain why the data points plotted this way fall close to a straight line. (b) Use your graph from part (a) to calculate the resistance \(R_{L}\) and inductance \(L\) of the solenoid. (c) If the current in the solenoid is \(20.0 \mathrm{~A}\), how much energy is stored there? At what rate is electrical energy being dissipated in the resistance of the solenoid?

A Radio Tuning Circuit. The minimum capacitance of a variable capacitor in a radio is \(4.18 \mathrm{pF}\). (a) What is the inductance of a coil connected to this capacitor if the oscillation frequency of the \(I\) - \(C\) circuit is \(1600 \times 10^{3} \mathrm{IIz}\), corresponding to one end of the AM radio broadcast band, when the capacitor is set to its minimum capacitance? (b) The frequency at the other end of the broadcast band is \(540 \times 10^{3} \mathrm{~Hz}\). What is the maximum capacitance of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?
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