Qwestrum Engineering360 · Electrical & Electronics · Electrical Machines – I
Testing of DC Machines
Testing methods estimate DC machine efficiency without full-load loading: the Swinburne test uses a no-load run to separate constant losses, while the Hopkinson (back-to-back) test loads two identical machines against each other.
Exam tip: keep SI units consistent end-to-end, write the governing relation symbolically before substituting, and sanity-check magnitude and sign.
Key formulas & points
Skim these first — then read the full notes below.
- Swinburne valid only for shunt machines at constant speed
- Retardation test for moment of inertia
- Efficiency at various loads from test data
Topic details
Introduction
The Swinburne test is a cheap indirect method: run the machine unloaded, measure input, and everything except a small friction/copper term is constant (core + friction + windage) loss. Copper loss at any load is then computed as I_a²R_a. Efficiency at any assumed load follows without physically loading the machine.
Scope in B.Tech and GATE syllabus
The Hopkinson test connects two identical machines mechanically coupled and electrically back-to-back; only the losses are drawn from the supply, so full-load conditions are reached with small input power — ideal for large machines.
Key relations & formulas
Formulas (Indian textbook notation)
Formulas (Indian textbook notation)
Formulas (Indian textbook notation)
Notation and sign conventions
Relation 1 —
Formulas (Indian textbook notation)
Write this relation with symbols exactly as in Electrical Machines — Nagrath & Kothari before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
Formulas (Indian textbook notation)
Write this relation with symbols exactly as in Electrical Machines — Nagrath & Kothari before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
Formulas (Indian textbook notation)
Write this relation with symbols exactly as in Electrical Machines — Nagrath & Kothari before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Fundamentals and definitions
In Swinburne, no-load input power minus no-load armature copper loss gives the constant loss W_c. For motoring efficiency at load current I_L: input = VI_L, total loss = W_c + I_a²R_a, and η = (input − loss)/input.
Governing relations in practice
The method assumes flux (hence core loss) stays constant, which is true only for shunt machines and fails when armature reaction is significant at heavy loads. It also cannot reveal commutation problems that appear only under load.
Design and analysis considerations
The brake test directly measures output torque via a spring balance and gives true efficiency for small machines: P_out = 2πNT/60.
Assumptions and validity limits
State assumptions explicitly before using any relation for testing of dc machines — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Electrical Machines I viva and GATE descriptive questions, listing valid assumptions often earns separate marks.
Step-by-step problem approach
1. Read the question and list given data with SI units (common in Electrical Machines I papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to testing of dc machines.
4. Use equation 1:
5. Use equation 2:
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to testing of dc machines.
4. Use equation 1:
.
5. Use equation 2:
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.
Applications & exam relevance
Testing of DC Machines appears in substations, drives, and labs. In Indian electrical curricula this topic is tested because it connects theory to magnetic circuits, transformers, and DC machines.
GATE and semester exams often combine testing of dc machines with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use testing of dc machines?" — answer with a lab, mini-project, or plant visit example if possible.
Common mistakes in exams
• Applying Swinburne to a series machine (flux is not constant)
• Treating the no-load input as pure constant loss without subtracting no-load armature copper loss
• Forgetting that copper loss scales with I_a² when projecting to full load
• Using rpm directly in 2πNT without the /60 conversion
• Treating the no-load input as pure constant loss without subtracting no-load armature copper loss
• Forgetting that copper loss scales with I_a² when projecting to full load
• Using rpm directly in 2πNT without the /60 conversion
Quick revision checklist
Before attempting testing of dc machines problems, confirm you can:
1. Swinburne valid only for shunt machines at constant speed
2. Retardation test for moment of inertia
3. Efficiency at various loads from test data
2. Retardation test for moment of inertia
3. Efficiency at various loads from test data
Revise the solved examples in Electrical Machines — Nagrath & Kothari and one previous-year GATE or university paper for this unit.
Worked examples
Try the problem first — open the solution when you are ready to check.
Efficiency by Swinburne test
Problem
A 250 V shunt motor takes 2 A at no load; R_a = 0.4 Ω, shunt field current 1 A. Estimate the efficiency when the line current is 40 A.
Solution
No-load armature current = 2 − 1 = 1 A; no-load Cu loss = 1²×0.4 = 0.4 W.
No-load input = 250×2 = 500 W → constant loss W_c = 500 − 0.4 = 499.6 W ≈ 500 W.
At line current 40 A: I_a = 40 − 1 = 39 A; armature Cu loss = 39²×0.4 = 608.4 W.
Field loss = 250×1 = 250 W; total loss = 500 + 608.4 + 250 = 1358 W.
Input = 250×40 = 10000 W; η = (10000 − 1358)/10000 = 86.4%.
No-load input = 250×2 = 500 W → constant loss W_c = 500 − 0.4 = 499.6 W ≈ 500 W.
At line current 40 A: I_a = 40 − 1 = 39 A; armature Cu loss = 39²×0.4 = 608.4 W.
Field loss = 250×1 = 250 W; total loss = 500 + 608.4 + 250 = 1358 W.
Input = 250×40 = 10000 W; η = (10000 − 1358)/10000 = 86.4%.
Conceptual check — Testing of DC Machines
Problem
In a Electrical Machines I semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of testing of dc machines." What should a complete answer include?
Exams & GATE
Nagrath & Kothari — Swinburne test numerical for efficiency curve.
📖 Standard books (India)
Electrical Machines — Nagrath & Kothari
Read: Syllabus unit
Transformers, DC machines, and induction motors
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