Problem 21 - Applied Physics

Question

The current in the secondary coil of a transformer is \(5.00 \mathrm{~A}\). Find the voltage in the secondary if the power is \(775 \mathrm{~W}\).

Step-by-Step Solution

Answer

The voltage in the secondary coil is 155 V.

Step 1: Understand the Relationship Between Power, Voltage, and Current

In electrical circuits, the power, voltage, and current are related by the formula \( P = VI \), where \( P \) is the power in watts, \( V \) is the voltage in volts, and \( I \) is the current in amperes. We will use this relationship to find the voltage in the secondary coil.

Step 2: Rearrange the Formula to Solve for Voltage

We need to find the voltage \( V \). To do this, rearrange the formula \( P = VI \) to solve for \( V \). By dividing both sides of the equation by \( I \), we have:\[ V = \frac{P}{I} \]

Step 3: Substitute the Given Values into the Equation

From the problem, we know that the power \( P \) is \( 775 \, \mathrm{W} \) and the current \( I \) is \( 5.00 \, \mathrm{A} \). Substitute these values into the equation to find the voltage:\[ V = \frac{775 \, \mathrm{W}}{5.00 \, \mathrm{A}} \]

Step 4: Calculate the Voltage in the Secondary Coil

Perform the division:\[ V = 155 \, \mathrm{V} \]This is the voltage in the secondary coil.

Key Concepts

Electrical Circuits

Understanding electrical circuits is a fundamental part of learning physics and electrical engineering. An electrical circuit consists of a continuous path in which electrons flow. This path is typically made of wires and is powered by a voltage source, like a battery or transformer.
In a basic electrical circuit, three main elements are critical: voltage, current, and power. Every circuit has these essential components:

• Voltage (V): The potential difference that drives electron flow. It's the "push" needed to get the current moving. Measured in volts (V).


• Current (I): The flow of electric charge within the circuit. Measured in amperes, or amps (A).


• Power (P): The rate at which electrical energy is consumed or converted. It's the product of voltage and current, measured in watts (W).

For transformer circuits, these principles remain consistent, and understanding them helps in solving transformer-related problems.

Power-Voltage-Current Relationship

In electrical systems, the relationship between power, voltage, and current is crucial. These quantities are interdependent and described by the formula: \[ P = VI \]Here, \( P \) represents power, \( V \) stands for voltage, and \( I \) symbolizes current. This equation is known as the Power Formula and is used to calculate any one of the three variables if the other two are known. It is a foundational concept in understanding how electrical circuits work.
When dealing with transformers, this relationship helps determine the characteristics of the output based on input parameters.
For example, to find the voltage in a secondary coil when the power and current are known, you manipulate the formula to solve for \( V \):\[ V = \frac{P}{I} \]
This rearrangement shows how the power divided by current gives the voltage, which helps in troubleshooting and designing circuits.

Secondary Coil Voltage

The voltage in the secondary coil of a transformer is a result of energy transfer between two circuits through electromagnetic induction. In transformers, the primary coil induces voltage into the secondary coil, allowing electric power to be transferred from one coil to another.
To find the secondary coil voltage, you first need to understand the principle of the transformer operation. The voltage in the secondary coil can be calculated if the power output and the current in the secondary coil are known. Using the power formula:
\[ V = \frac{P}{I} \]
Where

• \( P \) (Power): The power delivered by the transformer in watts.
• \( I \) (Current): The current flowing through the secondary coil in amperes.
• \( V \) (Voltage): The resulting voltage produced in the secondary coil in volts.

From the example problem, with a power of \( 775 \, \text{W} \) and a current of \( 5.00 \, \text{A} \), the voltage in the secondary coil is calculated as \( 155 \, \text{V} \). This calculation demonstrates how energy is efficiently transferred and transformed in electrical circuits.