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Chapter 22: Problem 61

mmh The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is \(50 : 1 .\) The primary coil is plugged into a standard \(120-\mathrm{V}\) outlet. The current in the secondary coil is \(1.7 \times 10^{-3}\) A. Find the power consumed by the air filter.

Short Answer

Expert verified
The power consumed by the air filter is 10.2 W.

Step by step solution

01

Identify the Given Information

We are given the following values:- Turns ratio of transformer: \( 50:1 \)- Primary voltage \( V_p = 120 \text{ V} \)- Current in the secondary coil \( I_s = 1.7 \times 10^{-3} \) A.
02

Use the Turns Ratio to Find Secondary Voltage

The turns ratio in the transformer allows us to determine the voltage in the secondary coil using the formula \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). Hence,\[V_s = V_p \times \frac{N_s}{N_p} = 120 \text{ V} \times 50 = 6000 \text{ V}.\]
03

Calculate the Power in the Secondary Coil

Power in the secondary coil is calculated using the formula \( P = V_s \times I_s \). Substitute the known values:\[P = 6000 \text{ V} \times 1.7 \times 10^{-3} \text{ A} = 10.2 \text{ W}.\]

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Key Concepts

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

Transformer
A transformer is an electrical device used to change the voltage levels between circuits. It operates on the principle of electromagnetic induction and typically consists of two coils: a primary coil and a secondary coil. These coils are wound around a common core.

When an alternating current flows through the primary coil, it creates a magnetic field. This magnetic field induces a voltage in the secondary coil. Depending on the coil winding, transformers can either step up (increase) or step down (decrease) the voltage.

Transformers are essential in the distribution of electrical energy, allowing for efficient transmission over long distances.
Turns Ratio
The turns ratio is a key property of a transformer and is defined as the ratio of the number of turns in the secondary coil (N_s) to the number of turns in the primary coil (N_p). It is represented as:
  • \( \text{Turns Ratio} = \frac{N_s}{N_p} \)
The turns ratio dictates how much the voltage will increase or decrease. In a step-up transformer, the secondary coil has more turns, which means the voltage is higher than what is applied to the primary coil.

In our example, the turns ratio is 50:1, which means the secondary coil has 50 times more turns than the primary.
Voltage
Voltage, in the context of transformers, refers to the electrical potential difference between two points. It is an important variable because it determines how much energy can be powered through a circuit.

Using the turns ratio, we can find the secondary voltage (V_s) from the primary voltage (V_p) using:
  • \( V_s = V_p \times \frac{N_s}{N_p} \)
In our specific case, a primary voltage of 120 V is transformed into a secondary voltage of 6000 V, thanks to the 50:1 turns ratio.
Current
Current is the flow of electric charge in a circuit and is measured in amperes (A). It plays a crucial role in determining the power that a device can consume.

In a transformer, the relationship between primary and secondary currents is inversely proportional to that of the voltage and turns ratio. This means if the voltage is increased, the current decreases, according to the equation:
  • \( \frac{I_s}{I_p} = \frac{N_p}{N_s} \)
Here, the current in the secondary coil is 1.7 脳 10^{-3} A, which is small, but is balanced by the high voltage it delivers.
Power Calculation
Power is the rate at which energy is consumed or transferred and is measured in watts (W). For transformers, power can be calculated using the formula:
  • \( P = V \times I \)
This tells us that power is the product of voltage and current. In a transformer, the power in the primary and secondary circuits can be considered almost equal, due to very small energy losses.

In our situation, the power consumed by the air filter is calculated as follows:\( P = 6000 \text{ V} \times 1.7 \times 10^{-3} \text{ A} = 10.2 \text{ W} \),
highlighting the efficiency with which transformers convert and deliver energy.

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Most popular questions from this chapter

Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surface is tilted at an angle of \(12^{\circ}\) from the horizontal, and a net magnetic flux of \(8.4 \times 10^{-3}\) Wb passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. (b) Find the smallest possible value for the area of the second surface.

Multiple-Concept Example 13 reviews the concepts used in this problem. A long solenoid (cross-sectional area \(=1.0 \times 10^{-6} \mathrm{m}^{2}\) , number of turns per unit length \(=2400\) turns/m) is bent into a circular shape so it looks like a donut. This wire-wound donut is called a toroid. Assume that the diameter of the solenoid is small compared to the radius of the toroid, which is 0.050 \(\mathrm{m}\) . Find the emf induced in the toroid when the current decreases to 1.1 A from 2.5 A in a time of 0.15 s.

At its normal operating speed, an electric fan motor draws only 15.0\(\%\) of the current it draws when it just begins to turn the fan blade. The fan is plugged into a \(120.0-\mathrm{V}\) socket. What back emf does the motor generate at its normal operating speed?

Coil 1 is a flat circular coil that has \(N_{1}\) turns and a radius \(R_{1}\) . At its center is a much smaller flat, circular coil that has \(N_{2}\) turns and radius \(R_{2}\) . The planes of the coils are parallel. Assume that coil 2 is so small that the magnetic field due to coil 1 has nearly the same value at all points covered by the area of coil \(2 .\) Determine an expression for the mutual inductance between these two coils in terms of \(\mu_{0}, N_{1}, R_{1}, N_{2},\) and \(R_{2}\) .

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?
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