# Lesson 9 Flashcards — Transformers and Power Transmission Click a question to reveal the answer.
1. Why does long-distance transmission run at high voltage?

Line loss is \(I^2R_\text{line}\). For a fixed delivered power \(P = VI\), raising \(V\) lowers \(I\), and the loss falls with the square of the current. Ten times the voltage means one hundredth the loss.

2. Why must the grid be AC rather than DC?

Because transformers require a changing magnetic flux. With DC the flux is constant, nothing is induced, and a transformer does nothing at all.

3. What physical law makes a transformer work?

Faraday's law — a changing magnetic flux through a coil induces a voltage across that coil. Two coils on a shared iron core see the same flux \(\phi\), and each coil's voltage is proportional to its own turn count: \(V_1 = N_1\,\partial\phi/\partial t\) and \(V_2 = N_2\,\partial\phi/\partial t\).

4. State the turns-ratio relationship.

\(a = N_1/N_2 = V_1/V_2 = I_2/I_1\). Note that current appears inverted: voltage and current move in opposite directions.

5. How do you tell a step-up transformer from a step-down one?

\(a < 1\) steps up (\(V_2 > V_1\)); \(a > 1\) steps down.

6. What does an ideal transformer conserve?

Power: \(P_1 = P_2\), so \(V_1I_1 = V_2I_2\). That is precisely why the current ratio inverts.

7. Which current do you use when computing line loss?

The current in that segment — the line current, not the load current. This is the single most common error on Quiz 3.

8. How do you compute line resistance from a per-kilometer figure?

\(R_\text{line} = r \times d\), where \(r\) is the conductor's resistance per unit length in Ω/km and \(d\) is its length. A typical overhead conductor runs \(r = 250\ \text{m}\Omega/\text{km}\). Note \(r\) is not \(\rho\): resistivity is a material property in \(\Omega\cdot\text{m}\), and dividing it by a cross-sectional area is what produces \(r\).

9. A 12 kW load at 120 V sits 20 km away with no transformer. What happens?

\(I = 100\ \text{A}\), \(R = 5\ \Omega\), so \(P_\text{line} = 50\ \text{kW}\) — over four times the delivered power, for \(12/62 = 19.35\%\) efficiency. Stepping up to 12 kV first drops the loss to 5 W.

10. What does \(V_\text{line}\) mean, and how does it get you back to the source voltage?

It is the drop across the wire, \(V_\text{line} = I\,R_\text{line}\) — never the voltage at a transformer terminal. KVL then walks you back to the source: \(V_s = V_1 + V_\text{line}\). In the Lesson 9 Your Turn the transformer's line side sits at 1200 V and the line drops \((1.25)(4) = 5\ \text{V}\), so \(V_s = 1200 + 5 = 1205\ \text{V}\). Read \(V_\text{line}\) as the 1200 V terminal voltage instead and KVL will not close.

11. What is the general method for a transmission system?

Work backward from the load: find load current, step through each transformer with \(a\), compute \(I^2R\) on the line, add it to get source power, then \(\eta = P_\text{load}/P_\text{source}\).