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T-04: Equivalent Circuit

ECE 2207 — Exam-Style Answers (Sorted by Topic)

Topic Overview: This document compiles all semester final questions and exam-style answers on Equivalent Circuit from 7 years of exams (2017–2024). Repeated questions appear once with all exam appearances noted.

[2017 Q8(c)]

📋 Appeared in: 2017 Q8(c)

(c) 200/400V step-up transformer, parameters referred to LV side: $R_{e q} = 0.15 Ω$ , $X_{e q} = 0.37 Ω$ , $R_{c} = 600 Ω$ , $X_{m} = 300 Ω$ . Load: 10A at 0.8 pf lag (secondary). Find: (i) primary current, (ii) secondary terminal voltage. [05]

Given: Turns ratio $a = N_{1} / N_{2} = 200/400 = 0.5$ (step-up), all parameters on LV (primary) side.

Load referred to primary side:

Secondary current $I_{2} = 10$ A. Referred to primary: $I_{2}^{'} = I_{2} / a = 10/0.5 = 20$ A (but we need to be careful: referred secondary current to primary = $I_{2} \times (N_{2} / N_{1}) = 10 \times 2 = 20$ A).

Wait: parameters are referred to LV side. Secondary current (HV side) = 10 A. Referred to LV (primary): $I_{2}^{'} = 10 \times (N_{2} / N_{1}) = 10 \times 2 = 20$ A at pf $= 0.8$ lag.

Taking $V_{2}^{'}$ as reference on primary side:

Secondary terminal voltage referred to primary: $V_{2}^{'} = 200$ V (at rated voltage the secondary is 400V, referred to primary = 400 × 0.5 = 200V).

Let $V_{2}^{'} = 200∠0°$ V, $I_{2}^{'} = 20∠ - 36.87°$ A

Approximate primary voltage (neglecting shunt branch for initial calc): $V_{1} = V_{2}^{'} + I_{2}^{'} (R_{e q} + j X_{e q})$ $= 200∠0° + 20∠ - 36.87° \times (0.15 + j 0.37)$

$I_{2}^{'} = 20 (0.8 - j 0.6) = 16 - j 12$

$I_{2}^{'} (R_{e q} + j X_{e q}) = (16 - j 12) (0.15 + j 0.37)$ $= 16 (0.15) + 16 (j 0.37) + (- j 12) (0.15) + (- j 12) (j 0.37)$ $= 2.4 + j 5.92 - j 1.8 + 4.44$ $= 6.84 + j 4.12$

$V_{1} = (200 + 6.84) + j 4.12 = 206.84 + j 4.12$ $∣ V_{1} ∣ = 206.8 4^{2} + 4.1 2^{2} \approx 206.88 V$

Primary current (including magnetizing branch):

$I_{c} = \frac{V _{1}}{R _{c}} = \frac{206.88}{600} = 0.345 A (in phase with V_{1})$ $I_{m} = \frac{V _{1}}{X _{m}} = \frac{206.88}{300} = 0.690 A (lagging V_{1} by 90°)$

No-load current: $I_{0} = I_{c} - j I_{m} = 0.345 - j 0.690$

$I_{1} = I_{0} + I_{2}^{'} = (0.345 - j 0.690) + (16 - j 12) = 16.345 - j 12.69$

$∣ I_{1} ∣ = 16.34 5^{2} + 12.6 9^{2} = 267.2 + 161.1 = 428.3 \approx 20.70 A$

Secondary terminal voltage (actual): Referred back to secondary side: $V_{2, actual} = ∣ V_{2}^{'} ∣ \times (N_{2} / N_{1}) = 200 \times 2 = 400 V$

(In this simplified case, since we set $V_{2}^{'} = 200$ V as reference, actual secondary $= 400$ V with the given load conditions.)

[2020 Q1(b)]

📋 Appeared in: 2020 Q1(b)

(b) Derive the equivalent circuit of a single-phase two-winding transformer. [04]

Step 1: Ideal transformer with no losses, no leakage: $\frac{V _{1}}{V _{2}} = \frac{N _{1}}{N _{2}} = a, I_{1} = \frac{I _{2}}{a}$

Step 2: Add core loss and magnetizing current (shunt branch): Primary draws no-load current $I_{0} = I_{c} + j I_{m}$ even at no load.

$I_{c}$ in phase with $V_{1}$ : represented by $R_{c} = V_{1} / I_{c}$ in shunt.
$I_{m}$ lags $V_{1}$ by 90°: represented by $X_{m} = V_{1} / I_{m}$ in shunt.

Step 3: Add primary winding resistance and leakage reactance: Series elements $R_{1}$ and $j X_{1}$ on the primary side.

Step 4: Refer secondary to primary: Replace $R_{2}$ , $j X_{2}$ with $a^{2} R_{2} = R_{2}^{'}$ , $j a^{2} X_{2} = j X_{2}^{'}$ on the primary side.

Step 5: Final approximate equivalent circuit (shunt branch at input):

$V_{1} \to [R_{1} + j X_{1} + R_{2}^{'} + j X_{2}^{'}] \to E_{1}$

Shunt branch ( $R_{c} ∥ j X_{m}$ ) connected across $V_{1}$ .

For simplicity, combine series elements: $R_{01} = R_{1} + R_{2}^{'}, X_{01} = X_{1} + X_{2}^{'}$

Approximate equivalent circuit of transformer referred to primary

[2020 Q3(c)]

📋 Appeared in: 2020 Q3(c)

(c) Step-by-step equivalent circuit of a transformer referred to primary side. [05]

1. Ideal Transformer Core Model: Start with an ideal core (zero winding resistance, zero leakage flux, infinite permeability, zero core loss). $E_{1} = a E_{2}, I_{1} = I_{2} / a where a = N_{1} / N_{2}$ Step 1: Ideal Transformer Model

2. Winding Resistances and Leakage Reactances: Add practical winding series resistance ( $R_{1}, R_{2}$ ) and leakage reactance ( $X_{1}, X_{2}$ ) on both sides. $V_{1} = E_{1} + I_{1} (R_{1} + j X_{1}), E_{2} = V_{2} + I_{2} (R_{2} + j X_{2})$ Step 2: Practical Windings with Resistance and Leakage Reactance

3. Core Excitation Shunt Branch: Add a parallel branch across $E_{1}$ to model core iron loss ( $R_{c}$ ) and magnetizing reactance ( $X_{m}$ ). The total no-load current is $I_{0} = I_{c} + I_{m}$ . Step 3: Complete Circuit Model with Physical Core Shunt Branch

4. Transferring Secondary to Primary (Exact Equivalent Circuit): To eliminate the ideal transformer, transfer secondary parameters to the primary using $a^{2}$ : $R_{2}^{'} = a^{2} R_{2}, X_{2}^{'} = a^{2} X_{2}, V_{2}^{'} = a V_{2}, I_{2}^{'} = I_{2} / a$ Step 4: Exact Equivalent Circuit

5. Approximate Equivalent Circuit: Since $I_{0}$ is small and the primary voltage drop $I_{0} (R_{1} + j X_{1})$ is negligible, move the shunt branch to the primary terminals. The series impedances combine to $R_{01} = R_{1} + R_{2}^{'}$ and $X_{01} = X_{1} + X_{2}^{'}$ . Step 5: Approximate Equivalent Circuit

[2024 Q4(c)]

📋 Appeared in: 2024 Q4(c)

(c) Obtain the equivalent circuit of a transformer referred to the primary side. [03, CO1]

Referring secondary to primary:

Replace all secondary quantities with primary-referred (primed) values: $R_{2}^{'} = a^{2} R_{2}, X_{2}^{'} = a^{2} X_{2}, E_{2}^{'} = a E_{2} = E_{1}, Z_{L}^{'} = a^{2} Z_{L}$

Final equivalent circuit referred to primary:

Series branch: $R_{01} = R_{1} + R_{2}^{'}$ , $X_{01} = X_{1} + X_{2}^{'}$ (total series impedance).

Shunt branch: $R_{c} ∥ j X_{m}$ (at primary terminals: approximate circuit).

Exact and approximate equivalent circuit referred to primary

In the approximate equivalent circuit, the shunt branch is moved to the primary input terminals (before $R_{1}$ , $X_{1}$ ). This simplifies calculation without significant error for most power transformers.

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6.7 KB · commit d2f3b6e

← T-03: No-Load & Phasors | 🏠 Index | T-05: Voltage Regulation →

T-04: Equivalent Circuit

ECE 2207 — Exam-Style Answers (Sorted by Topic)

Topic Overview: This document compiles all semester final questions and exam-style answers on Equivalent Circuit from 7 years of exams (2017–2024). Repeated questions appear once with all exam appearances noted.

[2017 Q8(c)]

📋 Appeared in: 2017 Q8(c)

Given: Turns ratio $a = N_{1} / N_{2} = 200/400 = 0.5$ (step-up), all parameters on LV (primary) side.

Load referred to primary side:

Secondary current $I_{2} = 10$ A. Referred to primary: $I_{2}^{'} = I_{2} / a = 10/0.5 = 20$ A (but we need to be careful: referred secondary current to primary = $I_{2} \times (N_{2} / N_{1}) = 10 \times 2 = 20$ A).

Wait: parameters are referred to LV side. Secondary current (HV side) = 10 A. Referred to LV (primary): $I_{2}^{'} = 10 \times (N_{2} / N_{1}) = 10 \times 2 = 20$ A at pf $= 0.8$ lag.

Taking $V_{2}^{'}$ as reference on primary side:

Secondary terminal voltage referred to primary: $V_{2}^{'} = 200$ V (at rated voltage the secondary is 400V, referred to primary = 400 × 0.5 = 200V).

Let $V_{2}^{'} = 200∠0°$ V, $I_{2}^{'} = 20∠ - 36.87°$ A

Approximate primary voltage (neglecting shunt branch for initial calc): $V_{1} = V_{2}^{'} + I_{2}^{'} (R_{e q} + j X_{e q})$ $= 200∠0° + 20∠ - 36.87° \times (0.15 + j 0.37)$

$I_{2}^{'} = 20 (0.8 - j 0.6) = 16 - j 12$

$I_{2}^{'} (R_{e q} + j X_{e q}) = (16 - j 12) (0.15 + j 0.37)$ $= 16 (0.15) + 16 (j 0.37) + (- j 12) (0.15) + (- j 12) (j 0.37)$ $= 2.4 + j 5.92 - j 1.8 + 4.44$ $= 6.84 + j 4.12$

$V_{1} = (200 + 6.84) + j 4.12 = 206.84 + j 4.12$ $∣ V_{1} ∣ = 206.8 4^{2} + 4.1 2^{2} \approx 206.88 V$

Primary current (including magnetizing branch):

$I_{c} = \frac{V _{1}}{R _{c}} = \frac{206.88}{600} = 0.345 A (in phase with V_{1})$ $I_{m} = \frac{V _{1}}{X _{m}} = \frac{206.88}{300} = 0.690 A (lagging V_{1} by 90°)$

No-load current: $I_{0} = I_{c} - j I_{m} = 0.345 - j 0.690$

$I_{1} = I_{0} + I_{2}^{'} = (0.345 - j 0.690) + (16 - j 12) = 16.345 - j 12.69$

$∣ I_{1} ∣ = 16.34 5^{2} + 12.6 9^{2} = 267.2 + 161.1 = 428.3 \approx 20.70 A$

Secondary terminal voltage (actual): Referred back to secondary side: $V_{2, actual} = ∣ V_{2}^{'} ∣ \times (N_{2} / N_{1}) = 200 \times 2 = 400 V$

(In this simplified case, since we set $V_{2}^{'} = 200$ V as reference, actual secondary $= 400$ V with the given load conditions.)

[2020 Q1(b)]

📋 Appeared in: 2020 Q1(b)

(b) Derive the equivalent circuit of a single-phase two-winding transformer. [04]

Step 1: Ideal transformer with no losses, no leakage: $\frac{V _{1}}{V _{2}} = \frac{N _{1}}{N _{2}} = a, I_{1} = \frac{I _{2}}{a}$

Step 2: Add core loss and magnetizing current (shunt branch): Primary draws no-load current $I_{0} = I_{c} + j I_{m}$ even at no load.

$I_{c}$ in phase with $V_{1}$ : represented by $R_{c} = V_{1} / I_{c}$ in shunt.
$I_{m}$ lags $V_{1}$ by 90°: represented by $X_{m} = V_{1} / I_{m}$ in shunt.

Step 3: Add primary winding resistance and leakage reactance: Series elements $R_{1}$ and $j X_{1}$ on the primary side.

Step 4: Refer secondary to primary: Replace $R_{2}$ , $j X_{2}$ with $a^{2} R_{2} = R_{2}^{'}$ , $j a^{2} X_{2} = j X_{2}^{'}$ on the primary side.

Step 5: Final approximate equivalent circuit (shunt branch at input):

$V_{1} \to [R_{1} + j X_{1} + R_{2}^{'} + j X_{2}^{'}] \to E_{1}$

Shunt branch ( $R_{c} ∥ j X_{m}$ ) connected across $V_{1}$ .

For simplicity, combine series elements: $R_{01} = R_{1} + R_{2}^{'}, X_{01} = X_{1} + X_{2}^{'}$

Approximate equivalent circuit of transformer referred to primary

[2020 Q3(c)]

📋 Appeared in: 2020 Q3(c)

(c) Step-by-step equivalent circuit of a transformer referred to primary side. [05]

[2024 Q4(c)]

📋 Appeared in: 2024 Q4(c)

(c) Obtain the equivalent circuit of a transformer referred to the primary side. [03, CO1]

Referring secondary to primary:

Replace all secondary quantities with primary-referred (primed) values: $R_{2}^{'} = a^{2} R_{2}, X_{2}^{'} = a^{2} X_{2}, E_{2}^{'} = a E_{2} = E_{1}, Z_{L}^{'} = a^{2} Z_{L}$

Final equivalent circuit referred to primary:

Series branch: $R_{01} = R_{1} + R_{2}^{'}$ , $X_{01} = X_{1} + X_{2}^{'}$ (total series impedance).

Shunt branch: $R_{c} ∥ j X_{m}$ (at primary terminals: approximate circuit).

Exact and approximate equivalent circuit referred to primary

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[2024 Q4(c)]

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[2020 Q1(b)]

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[2024 Q4(c)]

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