Question

A transformer with 1800 turns on the primary and 300 turns on the secondary is used in an electric slot car racing set to reduce the input voltage amplitude of \(170 \mathrm{V}\) from the wall output. The current in the secondary coil is of amplitude \(3.2 \mathrm{A}\). What is the voltage amplitude across the secondary coil and the current amplitude in the primary coil?

Short answer

Answer: The voltage amplitude across the secondary coil is approximately 28.33V, and the current amplitude in the primary coil is approximately 0.53A.

Step-by-step solution

Determine the Voltage-Turns Ratio

To calculate the voltage across the secondary coil, we first need to find the turns ratio between the primary and secondary coils. The turns ratio is the ratio of the number of turns on the primary coil to the number of turns on the secondary coil: $$ Turns \ Ratio = \frac{N_{primary}}{N_{secondary}} $$ With \(N_{primary} = 1800\) turns and \(N_{secondary} = 300\) turns: $$ Turns \ Ratio = \frac{1800}{300} = 6 $$

Calculate the Voltage Amplitude Across the Secondary Coil

Now, using the turns ratio, we can find the voltage amplitude across the secondary coil using the voltage-turns ratio formula: $$ V_{secondary} = \frac{V_{primary}}{Turns \ Ratio} $$ With \(V_{primary} = 170 \mathrm{V}\) and \(Turns \ Ratio = 6\): $$ V_{secondary} = \frac{170 \mathrm{V}}{6} \cong 28.33 \mathrm{V} $$ So, the voltage amplitude across the secondary coil is approximately \(28.33 \mathrm{V}\).

Calculate the Current-Turns Ratio

Now, let's determine the current-turns ratio, which is the inverse of the turns ratio: $$ Current \ Ratio = \frac{1}{Turns \ Ratio} = \frac{1}{6} $$

Calculate the Current Amplitude in the Primary Coil

We can now use the current-turns ratio to find the current amplitude in the primary coil. The current in the secondary coil is given as \(I_{secondary} = 3.2 \mathrm{A}\). Using the current-turns ratio formula: $$ I_{primary} = Current \ Ratio \times I_{secondary} $$ With \(Current \ Ratio = \frac{1}{6}\) and \(I_{secondary} = 3.2 \mathrm{A}\): $$ I_{primary} = \frac{1}{6} \times 3.2\mathrm{A} \cong 0.53 \mathrm{A} $$ So, the current amplitude in the primary coil is approximately \(0.53 \mathrm{A}\). In conclusion, the voltage amplitude across the secondary coil is approximately \(28.33 \mathrm{V}\), and the current amplitude in the primary coil is approximately \(0.53 \mathrm{A}\).