---
title: "230/2300 V transformer with load impedance"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-230-2300-load-impedance"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-230-2300-load-impedance
site: BCT Engineering
---

# 230/2300 V transformer with load impedance

Electrical Machines · Chapter 2: Transformer

- Asked in: 2070 Bhadra, 8 marks
- Video: [(E)EM Ex 2.2 || Equivalent Circuit of a Transformer || Refer to Pri & Secy sides ||Open-Short Test](https://www.youtube.com/watch?v=uJP_rFndPtU) — Electrical Engineering Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-230-2300-load-impedance
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

A 230 V / 2300 V single-phase transformer is excited by 230 V ac voltage. The equivalent resistance and reactance referred to primary side are $0.1\ \Omega$ and $0.4\ \Omega$ respectively. Given that $R_0 = 500\ \Omega$ and $X_0 = 200\ \Omega$. The load impedance is $(400 + j600)\ \Omega$. Calculate: (i) primary current and input power factor (ii) secondary terminal voltage.

## Short answer

Primary current is 31.478 A with input power factor 0.533 lagging. Secondary terminal voltage is 2182.1 V.

## Answer

Refer the load to the primary and use the approximate equivalent circuit, with the exciting branch ($R_0$, $X_0$) connected across the supply.

```text
        I'2   R01=0.1   X01=0.4
  o----->-----[ R ]-----[ X ]----+
                                 |
  V1 = 230 V   Io <- (R0 || X0)  [ Z'L = 4 + j6 ]
                  (across supply)|
  o------------------------------+
```

#### (i) Primary current and input power factor

Turns ratio: $a = 2300/230 = 10$ (the transformer steps up).

Load referred to primary:

$$
Z_L' = \frac{Z_L}{a^2} = \frac{400 + j600}{100} = 4 + j6\ \Omega
$$

Load branch impedance seen by the supply:

$$
Z = (R_{01} + jX_{01}) + Z_L' = (0.1 + j0.4) + (4 + j6) = 4.1 + j6.4 = 7.601\angle57.36^\circ\ \Omega
$$

Load current (referred to primary), taking $V_1 = 230\angle 0^\circ$:

$$
I_2' = \frac{230}{4.1 + j6.4} = 30.261\angle -57.36^\circ\ \text{A} = 16.323 - j25.480\ \text{A}
$$

No-load (exciting) current:

$$
\begin{aligned}
I_c &= \frac{230}{500} = 0.46\ \text{A}, \qquad I_m = \frac{230}{200} = 1.15\ \text{A} \\
I_0 &= 0.46 - j1.15\ \text{A}
\end{aligned}
$$

Primary current:

$$
I_1 = I_2' + I_0 = (16.323 - j25.480) + (0.46 - j1.15) = 16.783 - j26.630\ \text{A} = 31.478\angle -57.78^\circ\ \text{A}
$$

Input power factor:

$$
\cos\phi_1 = \cos(57.78^\circ) = 0.533\ \text{lagging}
$$

#### (ii) Secondary terminal voltage

Voltage across the referred load:

$$
V_2' = I_2' Z_L' = 30.261 \times |4 + j6| = 30.261 \times 7.211 = 218.21\ \text{V}
$$

$$
V_2 = a\,V_2' = 10 \times 218.21 = 2182.1\ \text{V}
$$

**Answer:** $I_1 = 31.478\ \text{A}$ at power factor $0.533$ lagging; secondary terminal voltage $= 2182.1\ \text{V}$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The 230/2300 V transformer has a = 10. The load (400 + j600) Ω referred to the primary is
   A. 4000 + j6000 Ω
   B. 40 + j60 Ω
   C. 4 + j6 Ω
   D. (400 + j600)/10 Ω
   Answer: C. Divide by a^2 = 100 when moving from secondary to primary: 4 + j6 Ω. (Key point: Referring the load)

2. In the approximate equivalent circuit used, the exciting branch R0 || X0 is connected
   A. After the series impedance, across the load
   B. In series with the load
   C. Across the supply, in parallel with the series branch plus load
   D. Not used at all
   Answer: C. The exciting current I0 = 0.46 - j1.15 A flows from the supply separately, so I1 = I2' + I0. (Key point: Exciting branch placement)

3. The load current I2' = 230/(4.1 + j6.4) = 30.26 A at -57.36°. How is the primary current obtained?
   A. I1 = I2' only
   B. I1 = I2' + I0 as a phasor sum, 31.48 A at 0.533 lagging
   C. I1 = I2' - I0
   D. I1 = I2' x 10
   Answer: B. Add the no-load current phasor: (16.323 - j25.480) + (0.46 - j1.15) = 16.783 - j26.630 A, magnitude 31.478 A. (Key point: Primary current)

4. The secondary terminal voltage is found by
   A. V2 = a x I2' x |Z'L| = 10 x 30.261 x 7.211 = 2182 V
   B. V2 = 230 V
   C. V2 = 230 x 10 = 2300 V
   D. V2 = I1 x |Z'L| x 10
   Answer: A. Voltage across the referred load is I2' x |4 + j6| = 218.2 V, which is multiplied by a = 10 to give 2182 V. (Key point: Secondary terminal voltage)

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