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Chapter 20: Problem 32

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?

Short Answer

Expert verified
Answer: The secondary current amplitude is 100 mA.

Step by step solution

01

Write down the relationship between the primary and secondary voltages and currents

For a transformer, the relationship between the primary and secondary voltages and currents is given by: $$\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}$$ Where \(V_p\) is the primary voltage, \(V_s\) is the secondary voltage, \(N_p\) is the number of turns in the primary coil, \(N_s\) is the number of turns in the secondary coil, \(I_p\) is the primary current, and \(I_s\) is the secondary current.
02

Determine the secondary voltage

We have the turns ratio \(N_p/N_s = 1/100\) and the primary voltage amplitude \(V_p = 170V\). To find the secondary voltage amplitude, we can use the relationship for voltages: $$\frac{V_p}{V_s} = \frac{N_p}{N_s}$$ Rearrange for \(V_s\), $$V_s = V_p \cdot \frac{N_s}{N_p}$$ Substitute the given values into the equation, $$V_s = 170V \cdot \frac{1}{100}$$ $$V_s = 1.7V$$
03

Determine the secondary current amplitude

Now, we know both the primary and secondary voltage amplitudes. We can use the relationship between the currents to find the secondary current amplitude. From the relationship equation, $$\frac{I_s}{I_p} = \frac{V_p}{V_s}$$ Rearrange the equation for \(I_s\), $$I_s = I_p \cdot \frac{V_p}{V_s}$$ Substitute the given values into the equation, $$I_s = 1.0mA \cdot \frac{170V}{1.7V}$$ $$I_s = 100 mA$$ The secondary current amplitude is \(100 mA\).

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Most popular questions from this chapter

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 \(B ?\)
A step-down transformer has 4000 turns on the primary and 200 turns on the secondary. If the primary voltage amplitude is \(2.2 \mathrm{kV},\) what is the secondary voltage amplitude?
A transformer for an answering machine takes an ac voltage of amplitude $170 \mathrm{V}\( as its input and supplies a \)7.8-\mathrm{V}$ amplitude to the answering machine. The primary has 300 turns. (a) How many turns does the secondary have? (b) When idle, the answering machine uses a maximum power of \(5.0 \mathrm{W}\). What is the amplitude of the current drawn from the 170 -V line?
A \(0.67 \mathrm{mH}\) inductor and a \(130 \Omega\) resistor are placed in series with a \(24 \mathrm{V}\) battery. (a) How long will it take for the current to reach \(67 \%\) of its maximum value? (b) What is the maximum energy stored in the inductor? (c) How long will it take for the energy stored in the inductor to reach \(67 \%\) of its maximum value? Comment on how this compares with the answer in part (a).
Calculate the equivalent inductance \(L_{\mathrm{eq}}\) of two ideal inductors, \(L_{1}\) and \(L_{2},\) connected in series in a circuit. Assume that their mutual inductance is negligible. [Hint: Imagine replacing the two inductors with a single equivalent inductor \(L_{\mathrm{cq}} .\) How is the emf in the series equivalent related to the emfs in the two inductors? What about the currents?]
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