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To solve this problem involving a transformer, we can use the transformer voltage ratio formula, which relates the number of turns in the coils to the voltages:

\[
\frac{V_s}{V_p} = \frac{N_s}{N_p}
\]

Where:
- \(V_s\) is the voltage on the secondary coil (what we are solving for),
- \(V_p = 240\,V\) is the voltage on the primary coil,
- \(N_s = 200\) is the number of turns on the secondary coil,
- \(N_p = 2000\) is the number of turns on the primary coil.

Rearranging this equation to solve for \(V_s\):

\[
V_s = V_p \times \frac{N_s}{N_p}
\]

Let's plug in the values to calculate \(V_s\).The voltage on the secondary coil of the transformer is \( 24 \, \text{V} \).

Would you like further details or have any questions?

Here are 5 related questions to consider:
1. What happens to the secondary voltage if the primary voltage is increased to 480 V?
2. How does changing the number of turns on the secondary coil affect the output voltage?
3. What would be the secondary current if the transformer is ideal and the load is known?
4. Can the transformer step up the voltage if the number of turns on the secondary coil exceeds the primary coil?
5. How does the efficiency of a transformer affect real-world voltage output?

**Tip:** In ideal transformers, power is conserved, meaning \( V_p \times I_p = V_s \times I_s \), where \( I \) represents current. This relationship helps when dealing with currents in transformer problems.

What is the voltage on the secondary coil of a transformer if there are 2000 turns on the primary coil and 200 on the secondary coil when there is 240 V on the primary coil?

Learn how to calculate the secondary voltage of a transformer using the transformer voltage ratio formula. With 2000 turns on the primary coil, 200 turns on the secondary coil, and 240V input, find out how to apply the correct formula for high school physics.

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