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(II) A coil has \(3.25 - \Omega\) resistance and \(440 - \mathrm { mH }\) inductance. If the current is 3.00\(\mathrm { A }\) and is increasing at a rate of 3.60\(\mathrm { A } / \mathrm { s }\) what is the potential difference across the coil at this moment?

Short Answer

Expert verified
The potential difference across the coil is 11.334 V.

Step by step solution

01

Understanding the Formula for Potential Difference Across a Coil

The potential difference across a coil, also known as the voltage, can be calculated using the formula: \( V = IR + L \frac{dI}{dt} \), where \( I \) is the current, \( R \) is the resistance, \( L \) is the inductance, and \( \frac{dI}{dt} \) is the rate of change of current.
02

Calculate the Voltage Due to Resistance

Calculate the voltage due to the resistance part using \( V_R = IR \). Given: \( I = 3.00 \) A, \( R = 3.25 \) \( \Omega \). Substitute these values into the formula: \( V_R = 3.00 \times 3.25 = 9.75 \) V.
03

Calculate the Voltage Due to Inductance

Calculate the voltage due to the inductance using \( V_L = L \frac{dI}{dt} \). Given: \( L = 440 \) mH \( = 0.440 \) H, \( \frac{dI}{dt} = 3.60 \) A/s. Substitute these values into the formula: \( V_L = 0.440 \times 3.60 = 1.584 \) V.
04

Sum the Voltages to Find the Total Potential Difference

The total potential difference across the coil is the sum of the voltage due to the resistance and the voltage due to the inductance: \( V = V_R + V_L = 9.75 + 1.584 = 11.334 \) V.

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Key Concepts

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

Resistance
Resistance in a coil refers to the opposition that the coil presents to the flow of electric current. It's measured in ohms (\( \Omega \)), a standard unit in electrical circuits. Think of resistance as a bottleneck in a hose; it restricts how easily current flows.
  • Higher resistance means less current flowing for the same voltage.
  • Lower resistance means more current flowing easily.
In our exercise, the coil has a resistance of \(3.25 \Omega\). With a given current of \(3.00 \text{ A}\), the voltage due to resistance is calculated by multiplying these two values: \( V_R = I \times R = 3.00 \times 3.25 \). This calculation results in \( V_R = 9.75 \text{ V}\). Understanding this is crucial because it shows how even a small resistance can lead to significant voltage requirements. Breaking it into pieces ensures you see how each part of the formula impacts the outcome.
Inductance
Inductance is another important property to understand when dealing with coils. It is a measure of the coil's ability to store energy in a magnetic field and resist changes in current flowing through it. Inductance is quantified in henrys (H).
  • A coil with high inductance will oppose rapid changes in current more strongly.
  • It is especially relevant in circuits with alternating current or rapidly changing current.
The inductance value, given as \(440 \text{ mH}\) or \(0.440 \text{ H}\), tells us how effectively the coil can react to changes in current. In the exercise, as the current changes at a rate of \(3.60 \text{ A/s}\), the voltage due to inductance, \( V_L\), is calculated using the formula: \( V_L = L \frac{dI}{dt} = 0.440 \times 3.60 \). This yields \( V_L = 1.584 \text{ V}\). Inductance demonstrates the dynamic relationship between the coil and the changing current.
Potential Difference
Potential difference, often referred to as voltage, represents the work needed to move a charge between two points. It's the sum of the voltage across resistance and voltage across inductance in this scenario. Potential difference can be thought of as the electrical "pressure" that drives current through a circuit.
Here the total potential difference across the coil is calculated by adding the voltages from both resistance and inductance: \( V = V_R + V_L \). Let's understand why these two are added:
  • The voltage due to resistance, \(9.75 \text{ V}\), accounts for the energy spent overcoming the coil's opposition to current flow.
  • The voltage from inductance, \(1.584 \text{ V}\), represents the energy necessary to counteract the change in magnetic field, as the current changes.
Adding these voltages shows the total potential difference required to maintain current as it changes through the coil. So the combined potential difference is \(11.334 \text{ V}\), helping us grasp how both resistance and inductance influence voltage in this electromagnetic context.

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Most popular questions from this chapter

(a) What is the rms current in a series \(L R\) circuit when a \(60.0-\mathrm{Hz}, 120-\mathrm{V}\) rms ac voltage is applied, where \(R=965 \Omega\) and \(L=225 \mathrm{mH} ? \quad\) (b) What is the phase angle between voltage and current? (c) How much power is dissipated? (d) What are the rms voltage readings across \(R\) and \(L ?\)

A \(35-\mathrm{mH}\) inductor with \(2.0-\Omega\) resistance is connected in series to a \(26-\mu \mathrm{F}\) capacitor and a \(60-\mathrm{Hz}, 45-\mathrm{V}(\mathrm{rms})\) source. Calculate \((a)\) the rms current, \((b)\) the phase angle, and \((c)\) the power dissipated in this circuit.

(a) What is the rms current in an \(R C\) circuit if \(R = 5.70 \mathrm { k } \Omega\) , \(C = 1.80 \mu \mathrm { F } ,\) and the rms applied voltage is 120\(\mathrm { V }\) at 60.0\(\mathrm { Hz }\) ? (b) What is the phase angle between voltage and current? (c) What is the power dissipated by the circuit? (d) What are the voltmeter readings across \(R\) and \(C ?\)

A 2200 -pF capacitor is charged to \(120 \mathrm{~V}\) and then quickly connected to an inductor. The frequency of oscillation is observed to be \(17 \mathrm{kHz}\). Determine \((a)\) the inductance, (b) the peak value of the current, and (c) the maximum energy stored in the magnetic field of the inductor.

Show that the fraction of electromagnetic energy lost (to thermal energy) per cycle in a lightly damped \(\left(R^{2} \ll 4 L / C\right) L R C\) circuit is approximately $$\frac{\Delta U}{U}=\frac{2 \pi R}{L \omega}=\frac{2 \pi}{Q}$$ The quantity \(Q\) can be defined as \(Q=L \omega / R,\) and is called the \(Q\) -value, or quality factor, of the circuit and is a measure of the damping present. A high \(Q\) -value means smaller damping and less energy input required to maintain oscillations.

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