--- title: "Transformer equivalent circuit referred to primary" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:equivalent-circuit-referred-primary" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/equivalent-circuit-referred-primary site: BCT Engineering --- # Transformer equivalent circuit referred to primary Electrical Machines I · Chapter 2: Transformer - Asked in: 2074 Bhadra, 8 marks - Video: [Equivalent Circuit of Transformer Referred to Primary and Secondary - Single Phase Transformer](https://www.youtube.com/watch?v=3Jc8MN8HiQI) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/equivalent-circuit-referred-primary - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question Draw the equivalent circuit of a transformer with their parameters as it is in primary side and secondary side. How all parameters can be transferred to primary side - explain with mathematical derivation. ## Short answer Secondary parameters are referred to primary by dividing voltage by K, multiplying current by K, and dividing impedances (R, X, Z) by K^2, where K=N2/N1. Equivalent series resistance R01 = R1 + R2/K^2 and reactance X01 = X1 + X2/K^2. Shunt branch remains unchanged. ## Answer The **equivalent circuit** of a transformer is a circuit of resistances and reactances that behaves exactly like the real transformer, so its performance can be calculated by simple circuit analysis. #### Exact equivalent circuit ```text I1 R1 X1 I2' R2' X2' o--->-/\/\--mmm--+--------+-->--/\/\--mmm--o + | I0 | + [ R0 ] [ X0 ] V1 E1 | | V2' - | | - o----------------+--------+----------------o primary side | secondary referred ``` Parameters: - $R_1, X_1$: primary winding resistance and leakage reactance. - $R_2, X_2$: secondary resistance and leakage reactance. - $R_0$: core-loss resistance (carries $I_w$, the working component of $I_0$). - $X_0$: magnetising reactance (carries $I_\mu$). In the actual transformer, $R_1, X_1, R_0, X_0$ are on the primary side and $R_2, X_2$ and the load $Z_L$ are on the secondary side, joined by an ideal transformer of ratio $K = N_2/N_1 = E_2/E_1$. #### Transferring secondary quantities to primary The rule: the referred quantity must give the **same power, losses and VA** as the original. **Voltage:** since $E_2 = KE_1$, $$ E_2' = \frac{E_2}{K} = E_1, \qquad V_2' = \frac{V_2}{K} $$ **Current:** by ampere-turn balance $N_1I_2' = N_2I_2$, $$ I_2' = KI_2 $$ **Resistance:** copper loss must be unchanged: $$ I_2'^2R_2' = I_2^2R_2 \;\Rightarrow\; R_2' = R_2\left(\frac{I_2}{I_2'}\right)^2 = \frac{R_2}{K^2} $$ **Reactance:** reactive power $I^2X$ must be unchanged: $$ I_2'^2X_2' = I_2^2X_2 \;\Rightarrow\; X_2' = \frac{X_2}{K^2} $$ **Load impedance:** similarly $Z_L' = \frac{Z_L}{K^2}$. Check with impedance: $Z_2' = \frac{V_2'}{I_2'} = \frac{V_2/K}{KI_2} = \frac{Z_2}{K^2}$. #### Approximate equivalent circuit referred to primary Since $I_0$ is only 2-5% of rated current, the shunt branch can be moved to the input terminals. Then the series parts combine: $$ \begin{aligned} R_{01} &= R_1 + R_2' = R_1 + \frac{R_2}{K^2} \\ X_{01} &= X_1 + X_2' = X_1 + \frac{X_2}{K^2} \\ Z_{01} &= \sqrt{R_{01}^2 + X_{01}^2} \end{aligned} $$ ```text o---+--------+----[ R01 ]----[ jX01 ]----o | | + [ R0 ] [ jX0 ] V2' = V2/K | | Z_L' = Z_L/K^2 o---+--------+---------------------------o ``` Similarly, referred to the secondary: $R_{02} = R_2 + K^2R_1$, $X_{02} = X_2 + K^2X_1$, and $R_0, X_0$ are multiplied by $K^2$. **Rule to remember:** going from the low-voltage side to the high-voltage side, impedances are multiplied by the square of the voltage ratio (HV/LV)$^2$; voltages scale by the ratio, currents by its inverse. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A secondary resistance R2 is referred to the primary (K = N2/N1) as: A. R2 K B. R2 / K C. R2 K^2 D. R2 / K^2 Answer: D. Copper loss must stay the same: I2'^2 R2' = I2^2 R2 with I2' = K I2, giving R2' = R2/K^2. (Key point: Resistance referral) 2. By ampere-turn balance, the secondary current referred to the primary is: A. I2' = I2 / K B. I2' = K I2 C. I2' = K^2 I2 D. I2' = I2 K / 2 Answer: B. N1 I2' = N2 I2 gives I2' = (N2/N1) I2 = K I2. (Key point: Current referral) 3. Why can the shunt branch (R0, X0) be moved to the input terminals in the approximate circuit? A. Because iron loss is zero B. Because the no-load current is only 2-5% of rated current C. Because R0 equals X0 D. Because the secondary is open Answer: B. I0 is small, so the voltage drop it causes in R1, X1 is negligible and the series parts can be combined as R01, X01. (Key point: Approximate circuit) 4. Referred to the primary, the total equivalent resistance is: A. R1 + R2 K^2 B. R1 + R2 / K C. R1 + R2 / K^2 D. R1 K^2 + R2 Answer: C. R01 = R1 + R2' = R1 + R2/K^2. 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