In an ideal transformer with a turns ratio of 2:1 (Np:Ns), the secondary voltage is what fraction of the primary voltage under ideal conditions?

In an ideal transformer with a turns ratio of 2:1 (Np:Ns), the secondary voltage is what fraction of the primary voltage under ideal conditions?

• A. Equal
• B. Twice
• C. Half
• D. One-third

Answer: (C)

In an ideal transformer, voltages scale with the turns ratio. The relationship is Vp/Vs = Np/Ns, so Vs = Vp × (Ns/Np). With a turns ratio of 2:1 (Np:Ns = 2:1), Ns/Np = 1/2, giving Vs = Vp × 1/2 = Vp/2. So the secondary voltage is half of the primary voltage under ideal conditions. For completeness, power is conserved in an ideal transformer, so Is = (Vp/Vs) × Ip. Since Vp/Vs = 2 here, Is = 2 × Ip, meaning the current on the secondary doubles when the voltage is halved. This all assumes perfect coupling and no losses.

In an ideal transformer, voltages scale with the turns ratio. The relationship is Vp/Vs = Np/Ns, so Vs = Vp × (Ns/Np). With a turns ratio of 2:1 (Np:Ns = 2:1), Ns/Np = 1/2, giving Vs = Vp × 1/2 = Vp/2. So the secondary voltage is half of the primary voltage under ideal conditions.

For completeness, power is conserved in an ideal transformer, so Is = (Vp/Vs) × Ip. Since Vp/Vs = 2 here, Is = 2 × Ip, meaning the current on the secondary doubles when the voltage is halved. This all assumes perfect coupling and no losses.