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ssm A step-down transformer (turns ratio \(=1 : 8 )\) is used with an electric train to reduce the voltage from the wall receptacle to a value needed to operate the train. When the train is running, the current in the secondary coil is 1.6 \(\mathrm{A}\) . What is the current in the primary coil?

Short Answer

Expert verified
The current in the primary coil is 0.2 A.

Step by step solution

01

Understand the problem

We have a step-down transformer with a turns ratio of 1:8, which means the primary coil has 8 times more turns than the secondary coil. We need to find the current in the primary coil given the current in the secondary coil is 1.6 A.
02

Recall the transformer current relationship

The transformer current relationship is given by the formula: \( I_p \cdot N_p = I_s \cdot N_s \), where \( I_p \) and \( I_s \) are the currents in the primary and secondary coils, respectively, and \( N_p \) and \( N_s \) are the numbers of turns. However, this can be simplified for the turns ratio as \( \frac{I_p}{I_s} = \frac{N_s}{N_p} \).
03

Apply the turns ratio to the current formula

Given the turns ratio \( 1:8 \), it means \( \frac{N_s}{N_p} = \frac{1}{8} \). Hence, \( I_p = I_s \times \frac{N_s}{N_p} = 1.6 \times \frac{1}{8} \).
04

Calculate the primary current

Substitute the given secondary current into the equation: \( I_p = 1.6 \times \frac{1}{8} = 0.2 \text{ A} \).

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Key Concepts

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

Turns Ratio
The concept of the turns ratio is fundamental in understanding how transformers work. A turns ratio is simply a way of comparing the number of turns in the primary coil to the number of turns in the secondary coil. In the given exercise, the turns ratio is 1:8. This means that for every turn in the secondary coil, the primary coil has eight turns.

This ratio directly affects how voltages are transformed. A step-down transformer, like in the problem, reduces voltage from a higher level to a lower level. The turns ratio also plays a crucial role in determining how current changes between the primary and secondary coils.
  • If the turns ratio is greater than 1, the transformer steps down voltage and steps up current.
  • A direct correlation exists: more turns in the primary means greater primary voltage and lesser primary current and vice versa for the secondary.
Understanding the turns ratio helps in predicting how a transformer will affect both voltage and current in an electrical circuit.
Primary Current
The primary current in a transformer is the current that flows through the primary coil. This current is pivotal to the transformer's operation, as it creates the magnetic field that enables the voltage transformation. In our example, we are asked to find this primary current.

The relationship between the primary and secondary current can be described using the formula:\[\frac{I_p}{I_s} = \frac{N_s}{N_p}\]where:
  • \( I_p \) is the primary current.
  • \( I_s \) is the secondary current.
  • \( N_s \) and \( N_p \) are the number of turns in the secondary and primary coils, respectively.
Using this equation, we can rearrange to solve for the primary current:\[I_p = I_s \times \frac{N_s}{N_p}\]Given that \( I_s = 1.6 \) A and \( \frac{N_s}{N_p} = \frac{1}{8} \), substituting these values, we find that:\[I_p = 1.6 \times \frac{1}{8} = 0.2 \text{ A}\]Thus, in this example, the primary current is 0.2 A. This illustrates how we can calculate the primary current when we know the turns ratio and the secondary current.
Step-down Transformer
A step-down transformer is designed to decrease the voltage from a primary side to a secondary side. It does this through electromagnetic induction and a higher number of turns in the primary coil compared to the secondary coil. This configuration reduces the voltage output and increases the current output in the secondary coil.

Step-down transformers are commonly used in various applications, including adapters for household electronics to reduce the mains voltage to the safer levels needed for devices. In the exercise, the step-down transformer reduces the wall outlet voltage for the electric train to operate safely.

The efficiency of the transformer is one of its biggest advantages. While stepping down the voltage, it compensates by increasing the current on the secondary circuit, illustrated by the transformation equations:\[V_p \cdot I_p = V_s \cdot I_s\]where \( V \) represents voltage and \( I \) represents current for the respective coils.
  • Primary Voltage \( V_p \) is reduced in the secondary circuit \( V_s \).
  • Primary Current \( I_p \) is increased to Secondary Current \( I_s \).
This principle ensures energy conservation across the system, keeping power consistent while adjusting voltage and current levels to meet the specific needs of connected devices.

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Most popular questions from this chapter

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes- through the square loop is \(7.0 \times 10^{-3}\) Wb. What is the flux that passes through the circular loop?

A uniform magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that an emf of 0.80 \(\mathrm{V}\) and a current of 3.2 \(\mathrm{A}\) are induced in the coil. The wire is then reformed into a single-turn square coil, which is used in the same magnetic field (again perpendicular to the plane of the coil and with a magnitude changing at the same rate). What emf and current are induced in the square coil?

The magnetic flux that passes through one turn of a 12 -turn coil of wire changes to 4.0 \(\mathrm{from} 9.0 \mathrm{Wb}\) in a time of 0.050 \(\mathrm{s}\) . The average induced current in the coil is 230 \(\mathrm{A}\) . What is the resistance of the wire?

A flat coil of wire has an area \(A, N\) turns, and a resistance \(R .\) It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of \(90^{\circ},\) so that the normal becomes perpendicular to the magnetic field. The coil has an area of \(1.5 \times 10^{-3} \mathrm{m}^{2}, 50\) turns, and a resistance of 140\(\Omega .\) During the time while it is rotating, a charge of \(8.5 \times 10^{-5}\) C flows in the coil. What is the magnitude of the magnetic field?

ssm Multiple-Concept Example 13 reviews some of the principles used in this problem. Suppose you wish to make a solenoid whose self-inductance is 1.4 \(\mathrm{mH}\) . The inductor is to have a cross-sectional area of \(1.2 \times 10^{-3} \mathrm{m}^{2}\) and a length of 0.052 \(\mathrm{m}\) . How many turns of wire are needed?

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