Question

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?

Short answer

Answer: The number of turns in the secondary coil is approximately 14, and the amplitude of the current drawn from the 170V line is approximately 0.0294 A.

Step-by-step solution

Calculate the turns ratio of the transformer

The turns ratio of a transformer is given by the equation: \( turns \ ratio = \frac{V_{secondary}}{V_{primary}} = \frac{N_{secondary}}{N_{primary}} \) We can solve for the number of turns in the secondary coil: \( N_{secondary} = turns \ ratio \times N_{primary} \)

Calculate the number of turns in the secondary coil

We can now plug in the given values to find the number of turns in the secondary coil: \( N_{secondary} = \frac{7.8 \ V}{170 \ V} \times 300 \) \( N_{secondary} \approx 13.71 \) Since the number of turns must be a whole number, we can round it to the nearest integer: \( N_{secondary} \approx 14 \)

Calculate the amplitude of the current drawn from the 170V line

We know that the answering machine uses a maximum power of 5.0 W when idle. We can use this power and the primary voltage to find the current amplitude: \( P = V_{primary} \times I_{primary} \) Rearranging the equation to solve for the primary current amplitude, we get: \( I_{primary} = \frac{P}{V_{primary}} \) Now plug in the given values: \( I_{primary} = \frac{5.0 \ W}{170 \ V} \) \( I_{primary} \approx 0.0294 \ A \) Thus, the amplitude of the current drawn from the 170V line is approximately 0.0294 A.