Problem 42
Question
(I) A motor has an armature resistance of \(3.05 \Omega\). If it draws 7.20 A when running at full speed and connected to a 120 -V line, how large is the back emf?
Step-by-Step Solution
Verified Answer
The back EMF is 98.04 V.
1Step 1: Understanding Ohm's Law
Ohm's Law relates voltage (V), current (I), and resistance (R) with the formula \( V = IR \). In this problem, we'll use it to find voltage drop across the armature.
2Step 2: Calculate Voltage Drop Across Armature
Use the formula for Ohm's Law to calculate the voltage drop across the armature. With resistance \( R = 3.05 \Omega \) and current \( I = 7.20 \, A \), the voltage drop \( V_{drop} = I \times R = 7.20 \times 3.05 = 21.96 \, V \).
3Step 3: Determine Back EMF
The back EMF (Electromotive Force) is the difference between the supply voltage and the voltage drop across the armature. Therefore, \( \text{Back EMF} = 120 \, V - 21.96 \, V = 98.04 \, V \).
4Step 4: Verify Results
Ensure that the calculated back EMF makes sense in the context of the problem, which it does, since it is less than the supply voltage and accounts for the resistive losses.
Key Concepts
Electromotive Force (emf)Armature ResistanceVoltage Drop
Electromotive Force (emf)
Electromotive force, often abbreviated as emf, is a crucial concept in electric circuits. It refers to the energy provided by a cell or a battery to each coulomb of charge passing through the circuit. Think of emf as the driving force that pushes electrons around the circuit, allowing electric current to flow. It's measured in volts, just like regular voltage. However, there's a subtle difference.
- Emf vs Voltage: While voltage refers to the potential difference between two points in a circuit, emf is the potential difference across the terminals of a cell or battery when no current is flowing through it.
- Back Emf: This term comes into play, especially in motors. When a motor runs, it generates its own voltage called back emf. This happens because of the motor's spinning armature in a magnetic field. Back emf acts against the supply voltage to oppose the current's flow, limiting the current in the motor to a safe value.
Armature Resistance
Armature resistance is the resistance offered by the winding of a motor's armature. Every time current flows through the armature, it faces resistance—just like water running through a hose faces resistance from the sides of the hose. This resistance is invariably one of the main factors that affect a motor's operation.
- Significance: Knowing the armature resistance is essential for designing efficient motors and for calculating how much energy is lost as heat due to resistance.
- Calculation: Ohm's Law is typically used to determine how much voltage is dropped across the armature due to its resistance. In the problem, we calculated voltage drop using the formula \( V = IR \), where \( I \) is the current and \( R \) is the armature resistance.
- Impact: A higher armature resistance leads to higher energy losses and less efficiency, while a lower resistance results in less heat generation and more efficient motor operation.
Voltage Drop
Voltage drop is an essential concept related to the loss of energy in any working electrical circuit. As electric current moves through components like resistors, some electrical energy is transformed into heat and thus "drops" along the way. The voltage drop in a motor's armature can significantly impact the working performance of the motor.
- Calculation: To find the voltage drop across a component, use Ohm's Law: \( V_{drop} = I \times R \). This formula helped in our original solution, where current \( I \) flowed through the armature with resistance \( R \).
- Interplay with Ohm's Law: Ohm's Law is about balance. It ensures that the sum of all voltage drops in a circuit equals the total supplied voltage. In motors, the sum of the voltage drop across the armature and the back emf equals the supplied emf from the power source.
- Impact: Significant voltage drops mean more energy dissipated as heat, which can decrease efficiency and increase the chances of overheating the motor's parts.
Other exercises in this chapter
Problem 40
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