ENERGY IN TRANSFORMERS
4 marks · Normal| Transformer | Vp (V) | Ip (A) | Vs (V) | Is (A) |
|---|---|---|---|---|
| X | 120 | 0.25 | 12 | |
| Y | 240 | 0.15 | 48 | |
| Z | 110 | 1.0 | 22 |
Complete the table by calculating the missing output current (Is) for each transformer.
Explain briefly how you calculated the output current for transformer X.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) — · must show X: 2.5 A, Y: 0.75 A, Z: 5.0 A
(b) Used input power = output power (VpIp = VsIs), substitute values and solve for Is. · must show States power conservation or VpIp = VsIs
Common mistakes: Trying to use the turns or voltage ratio instead of power for current
Show mark scheme and worked solution
Marking points
A1 (a) · 1 mark · Correct value for X: Is = 2.5 A.
A2 (a) · 1 mark · Correct value for Y: Is = 0.75 A.
A3 (a) · 1 mark · Correct value for Z: Is = 5.0 A.
B1 (b) · 1 mark · Describes method as using input power = output power / VpIp = VsIs.
Worked solution
| Transformer | Vp (V) | Ip (A) | Vs (V) | Is (A) |
|---|---|---|---|---|
| X | 120 | 0.25 | 12 | 2.5 |
| Y | 240 | 0.15 | 48 | 0.75 |
| Z | 110 | 1.0 | 22 | 5.0 |
For each transformer: Is = (Vp × Ip) / Vs.
X: Is = (120 × 0.25) / 12 = 2.5 A. Y: Is = (240 × 0.15) / 48 = 0.75 A. Z: Is = (110 × 1.0) / 22 = 5.0 A.