current ratio

current ratio

1. What Current Ratio Is

Current ratio in ideal transformer is inverse of voltage/turns ratio:

(ideal, power in = power out: $V_1 I_1 = V_2 I_2$ for unity pf, or complex $V_1 I_1* = V_2 I_2*$).

So step-up voltage ($N_2>N_1$) ⇒ step-down current ($I_2<I_1$) and vice versa — conservation of power (minus losses).

For practical, $I_1 = I_0 + (N_2/N_1) I_2$ where $I_0$ is magnetizing current (small, ~2–5% of $I_1$ at full load) due to core.

2. Relation to Turns

If $a= V_1/V_2$, then $I_1 = I_2 /a$. This is used for reflected current.

3. Worked Examples

(a) Ideal N1=100, N2=1000 (a=0.1 step-up 10× voltage), I2=10A ⇒ I1=100A ✓ Power: V1·I1 =230·100=23kW, V2=2300·10=23kW.

(b) Practical with I0=2A (magnetizing) and I2=10A, a=0.1 ⇒ I1= I0 + a·I2? Actually I1= I0 + (N2/N1)·I2? No I1= I0 + (N2/N1)·I2? Wait ideal I1_ideal= (N2/N1)·I2=10·10=100A, plus I0 vectorially (90° lag) ⇒ |I1|≈100.02A.

(c) Need to find primary current for 400V/230V step-down 1kVA load: I2=4.35A, I1= (230/400)·4.35≈2.5A.

(d) Why high voltage low current for transmission? P=VI constant ⇒ higher V lower I ⇒ lower I²R loss.

4. Exam Points

• State $I_2/I_1=N_1/N_2$ (inverse) — 3-5 marks, very frequent (often paired with voltage ratio).
• Show power conservation $V_1 I_1≈V_2 I_2$ — 3 marks, frequent.
• Mention practical $I_1 = I_0 + (N_2/N_1)I_2$ — 5 marks, frequent.
• Solve $I_1$ for given $N$, $I_2$ — 3-5 marks, frequent.