Problem 9
Question
A \(11 \overline{0}-\Omega\) resistance coil draws a current of \(5.00 \mathrm{~A}\). What power is used?
Step-by-Step Solution
Verified Answer
The power used is 275.0 W.
1Step 1: Understanding the Formula for Power
To find the power used by the coil, we use the formula for electrical power: \( P = I^2 R \), where \( P \) is the power in watts, \( I \) is the current in amperes, and \( R \) is the resistance in ohms.
2Step 2: Substitute Known Values
We know from the problem statement that the resistance \( R \) is \( 11.0 \underline{\phantom{xxx}} \Omega \) and the current \( I \) is \( 5.00 \underline{\phantom{xxx}} A \). Substitute these values into the formula: \( P = (5.00)^2 \times 11.0 \).
3Step 3: Calculate the Power
First, square the current: \( (5.00)^2 = 25.00 \). Next, multiply by the resistance: \( 25.00 \times 11.0 = 275.0 \). The power used by the resistance coil is \( 275.0 \underline{\phantom{xxx}} W \).
Key Concepts
Ohm's LawResistanceCurrent CalculationPower Formula
Ohm's Law
Ohm's Law is a fundamental concept in electrical engineering and physics. It establishes a relationship between voltage, current, and resistance in an electrical circuit. According to Ohm's Law, the voltage (\( V \)) across a conductor is directly proportional to the current (\( I \)) flowing through it, with the proportionality constant being the resistance (\( R \)) of the conductor. Mathematically, this relationship is expressed as:\[ V = I \times R \]
- Voltage (\( V \)): measured in volts (V), it is the electrical potential difference between two points.
- Current (\( I \)): measured in amperes (A), it refers to the flow of electric charge through a conductor.
- Resistance (\( R \)): measured in ohms (Ω), it represents the opposition to the flow of electric current.
Resistance
Resistance is a measure of how much a material opposes the flow of electric current. It acts like friction in an electrical context. Materials have different levels of resistance:
Resistance in electrical components like resistors is always a consideration in designing circuits, as it impacts the efficiency and safety of the electrical system.
- Conductors: Low resistance materials like copper and aluminum allow electricity to flow easily.
- Insulators: High resistance materials such as rubber and glass block the flow of electricity.
Resistance in electrical components like resistors is always a consideration in designing circuits, as it impacts the efficiency and safety of the electrical system.
Current Calculation
Current calculation involves determining the amount of electric charge flowing through a circuit over a given period. Current is denoted by \( I \) and measured in amperes (A).To calculate current using Ohm's Law, you can rearrange the formula to solve for \( I \):\[ I = \frac{V}{R} \] When calculating current:- Ensure you have the voltage across the circuit component and the resistance of the component.- Substitute these values into the formula to find the current.For example, if you have a voltage of 10 volts across a resistor with a resistance of 2 ohms, the current through the resistor would be:\[ I = \frac{10 \, \text{V}}{2 \, \Omega} = 5 \, \text{A} \]Current calculation is essential in designing and troubleshooting circuits to ensure they work efficiently without overheating or damage.
Power Formula
The power formula is used to calculate the rate at which electrical energy is consumed or generated in a circuit. The expression for electrical power \( P \) is given by multiplying the current \( I \) by the voltage \( V \):\[ P = I \times V \]However, when dealing with resistance, another useful version is derived from Ohm's Law, expressed as:\[ P = I^2 \times R \]This formula is particularly helpful when you know the current flowing through a component and its resistance. It shows that power increases with both the square of the current and the resistance.In practical terms, electrical power is measured in watts (W), and calculating power is crucial to:- Determine the electrical load on components.- Ensure devices are used within their safe operating limits.- Manage the energy consumption in electrical systems.By understanding the power formula, you can efficiently design circuits and ensure they meet the necessary power requirements without overloading components.
Other exercises in this chapter
Problem 9
Find the capacitive reactance of a \(33 \overline{0}-\mu \mathrm{F}\) capacitor in a circuit of frequency \(30 \overline{0} \mathrm{~Hz}\).
View solution Problem 9
Find the current (in amperes) in each inductive circuit. \(L=5.00 \mathrm{mH}, f=2.00 \mathrm{kHz}, E=50.0 \mathrm{~V}\)
View solution Problem 10
Find the apparent power produced by a generating station whose actual power is \(1,250,000 \mathrm{~kW}\) and whose power factor is \(0.820\).
View solution Problem 10
Find the resonant frequency of a circuit containing a \(3.75-\mu \mathrm{F}\) capacitor in series with a \(30 \overline{0}-\mu \mathrm{H}\) inductor.
View solution