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Chapter 20: Problem 35

When the emf for the primary of a transformer is of amplitude $5.00 \mathrm{V},\( the secondary emf is \)10.0 \mathrm{V}$ in amplitude. What is the transformer turns ratio \(\left(N_{2} / N_{1}\right) ?\)

Short Answer

Expert verified
Answer: The transformer turns ratio (N2/N1) is 2.

Step by step solution

01

Understand the Transformer Turns Ratio Formula

The formula to relate the primary and secondary voltages of a transformer with its turns ratio is given by: \[ \frac{V_{2}}{V_{1}} = \frac{N_{2}}{N_{1}} \] where V1 is the primary voltage, V2 is the secondary voltage, N1 is the number of turns in the primary coil, and N2 is the number of turns in the secondary coil.
02

Plug in the given values

We are given V1 = 5.00 V and V2 = 10.0 V in the problem. We will plug these values into the formula: \[ \frac{10.0}{5.00} = \frac{N_{2}}{N_{1}} \]
03

Solve for N2/N1

Now, we will solve for the transformer turns ratio (N2/N1): \[ \frac{N_{2}}{N_{1}} = \frac{10.0}{5.00} = 2 \] So the transformer turns ratio N2/N1 is 2.

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Most popular questions from this chapter

A horizontal desk surface measures \(1.3 \mathrm{m} \times 1.0 \mathrm{m} .\) If Earth's magnetic field has magnitude \(0.44 \mathrm{mT}\) and is directed \(65^{\circ}\) below the horizontal, what is the magnetic flux through the desk surface?
A step-down transformer has a turns ratio of \(1 / 100 .\) An ac voltage of amplitude \(170 \mathrm{V}\) is applied to the primary. If the primary current amplitude is \(1.0 \mathrm{mA},\) what is the secondary current amplitude?
Calculate the equivalent inductance \(L_{\mathrm{eq}}\) of two ideal inductors, \(L_{1}\) and \(L_{2},\) connected in parallel in a circuit. Assume that their mutual inductance is negligible. [Hint: Imagine replacing the two inductors with a single equivalent inductor \(L_{\mathrm{eq}} .\) How is the emf in the parallel equivalent related to the emfs in the two inductors? What about the currents? \(]\)
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A flip coil is a device used to measure a magnetic field. A coil of radius $r, N\( turns, and electrical resistance \)R$ is initially perpendicular to a magnetic field of magnitude B. The coil is connected to a special kind of galvanometer that measures the total charge \(Q\) that flows through it. To measure the field, the flip coil is rapidly flipped upside down. (a) What is the change in magnetic flux through the coil in one flip? (b) If the time interval during which the coil is flipped is \(\Delta t,\) what is the average induced emf in the coil? (c) What is the average current that flows through the galvanometer? (d) What is the total charge \(Q\) in terms of \(r, N, R,\) and \(B ?\)
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