Integrated circuit and machine problem workshop

By the end of EE11001 you know several formulae that look unrelated: a PMMC ammeter's shunt equation, a DC motor's back-EMF relation, a wattmeter's power reading, a transformer's efficiency formula. Each was proved in its own chapter, with its own diagram, and none of them individually is new here. The workshop's whole point is that these formulae are not islands — an instrument reading is always somebody else's circuit or machine variable wearing a dial. A PMMC ammeter reading is the armature current $I_a$ that the torque and back-EMF equations expect. A wattmeter reading on an open-circuit test is the core loss $P_i$ that the efficiency formula expects. Nothing is invented; two formulae from different chapters are simply used back to back on the same physical quantity, in the order the physical setup dictates. This article rebuilds that chaining logic explicitly, works one instrument-to-machine problem and one instrument-to-transformer problem in full, and collects the traps that catch students who plug in a number without asking which symbol it actually is.

The core insight

An instrument reading is a specific circuit or machine symbol in disguise, so an integrated problem is solved by identifying which symbol each reading represents and then substituting it, unchanged, into the already-known formula from an earlier chapter.

The core idea

Three recurring instrument-meets-theory chains. Almost every integrated problem in this workshop is one of the following patterns, run in sequence rather than as one new law.

Chain 1 — current measurement into a machine equation. A PMMC ammeter with a shunt measures a DC current $I$ using

R_{sh} = \frac{I_m R_m}{I - I_m}

where: $R_{sh}$ is the shunt resistance, $I_m$ and $R_m$ are the meter coil's full-scale current and resistance, and $I$ is the full-scale current the shunted ammeter is designed to read. The value the ammeter displays while in service, once built, is then substituted directly as the armature current $I_a$ in a DC-motor formula such as the back-EMF relation

E_b = V - I_aR_a

where: $E_b$ is the back-EMF, $V$ is the supply voltage, and $R_a$ is the armature resistance.

Chain 2 — voltage and power readings into a power-factor calculation. A wattmeter reads active power $P = VI\cos\theta$ while a voltmeter and ammeter separately read the RMS voltage $V$ and current $I$ of the same load. Because all three instruments are reading the same load simultaneously, the load's power factor is recovered as

\cos\theta = \frac{P}{VI}

Chain 3 — OC/SC test readings into transformer efficiency. The wattmeter reading during an open-circuit (OC) test is the core loss $P_i$ ; the wattmeter reading during a short-circuit (SC) test at rated current is the copper loss $P_{cu}$ . Both feed the efficiency formula from the Transformers chapter, worked in the derivation below.

Figure

Three arrows show an instrument reading feeding as an input into a machine or circuit formula from an earlier chapter

Derivation AWhy OC and SC readings fix transformer efficiency

Show that the two wattmeter readings taken during the standard OC and SC tests are exactly the two loss terms the efficiency formula $\eta = VI\cos\theta/(VI\cos\theta+P_i+P_{cu})$ already expects, so no new measurement or law is required beyond the two test readings and the rated-load output.

Step 1. In the open-circuit test, the transformer's secondary is left open, so no current flows in the load-carrying winding and the primary draws only the small magnetising current needed to set up the core flux. Because $I$ in the untested winding is essentially zero, the only power the wattmeter can register is the power dissipated in the iron core by hysteresis and eddy currents — the winding's own $I^2R$ loss vanishes since there is no load current in that side. The wattmeter reading is therefore, by elimination, the core (iron) loss: $P_i = P_{OC}$ .

Step 2. In the short-circuit test, the secondary is short-circuited and only a small voltage is applied to the primary — just enough to circulate rated current through both windings. At this reduced voltage the core flux, and hence the core loss, is proportional to voltage squared and so is negligible compared to its rated-voltage value. The wattmeter reading under this condition is dominated by the $I^2R$ heating of the windings carrying rated current, so it directly gives the copper loss at rated current: $P_{cu} = P_{SC}$ .

Step 3. The transformer's efficiency at rated output is, by definition, output power divided by output power plus total losses, and total losses at rated load are exactly the sum of core loss (present whenever the transformer is energised, independent of load current) and copper loss (present only when current flows, and equal to its rated-current value here):

\eta = \frac{VI\cos\theta}{VI\cos\theta + P_i + P_{cu}}

Step 4. Substituting the identifications from Steps 1 and 2, $P_i = P_{OC}$ and $P_{cu} = P_{SC}$ , turns the definition into a formula built entirely from two wattmeter numbers and the rated output — which is exactly why the OC/SC test pair, and no further measurement, suffices to predict efficiency at any load.

Worked example

A single-phase transformer is tested exactly as in the syllabus procedure. On open circuit, a dynamometer wattmeter reads 150 W, a moving-iron voltmeter reads 230 V, and a moving-iron ammeter reads 1.2 A on the low-voltage side. On short circuit at rated current, the wattmeter reads 180 W. At its rated load the transformer delivers active power $VI\cos\theta = 3000\ \text{W}$ to a load at $\cos\theta = 0.9$ lagging. Find the efficiency at this load.

Step 1. Identify each reading against the derivation above: the OC wattmeter reading is the core loss, $P_i = 150\ \text{W}$ ; the SC wattmeter reading is the copper loss at rated current, $P_{cu} = 180\ \text{W}$ . The OC voltmeter and ammeter readings (230 V, 1.2 A) merely confirm the test was run at rated LV-side voltage and are not needed in the efficiency formula itself.

Step 2. Substitute into the efficiency formula proved above:

\eta = \frac{VI\cos\theta}{VI\cos\theta+P_i+P_{cu}} = \frac{3000}{3000+150+180} = \frac{3000}{3330}

Step 3. Evaluate the fraction.

\eta \approx 0.901

The transformer's efficiency at this load is approximately 90.1 percent.

Competitive stretch

Suppose instead the same PMMC ammeter design from the shunt formula ( $R_m = 2\ \Omega$ , $I_m = 2\ \text{mA}$ , built to read 0–20 A) is fitted to a shunt DC motor's armature line and indicates 15 A, with supply $V = 220\ \text{V}$ and armature resistance $R_a = 0.5\ \Omega$ . Chain 1 applies directly: the shunt itself is $R_{sh} = I_mR_m/(I-I_m) = (2\times10^{-3})(2)/(20-2\times10^{-3}) \approx 2.0\times10^{-4}\ \Omega$ , and using the reading $I_a = 15\ \text{A}$ in the back-EMF relation gives $E_b = 220 - (15)(0.5) = 212.5\ \text{V}$ . Two independent formulae, one shared number.

Common traps

Treating an instrument reading as a bare number rather than asking which symbol in which earlier formula it stands for ( $I_a$ ? $P_i$ ? $\cos\theta$ ?) before substituting it anywhere.
Swapping the OC and SC wattmeter roles — the OC reading is always core loss, the SC reading is always copper loss at the tested current; reversing them inverts the efficiency answer.
Using a reading from the wrong side of a transformer (primary vs. secondary) or the wrong branch of a machine (armature vs. line current) without checking which side the target formula expects.
Re-deriving a relation "from scratch" inside a workshop problem instead of recognising it as a direct, unmodified application of a formula proved earlier in the course.

Ask the tutor: if you are stuck deciding which earlier-chapter formula a given instrument reading belongs in, ask the AI tutor to trace the reading back to its symbol with you, step by step.

Why do OC and SC tests use different applied voltages instead of just running both at rated voltage

The OC test is run at rated voltage precisely because core loss depends on the working flux, so you need rated voltage to reproduce the true core loss. The SC test uses only a small applied voltage because its purpose is to push rated current through the windings without needing much voltage at all — with the secondary shorted, only the winding impedance limits the current. Running SC at rated voltage would drive dangerously large currents and tell you nothing extra about copper loss, which only needs rated current, not rated voltage.

If the ammeter shunt formula and the motor back-EMF formula come from completely different chapters, why is chaining them together considered a new type of problem

The two formulae are individually old, but recognising that the ammeter's displayed number is exactly the $I_a$ the machine formula wants is the actual skill being tested. Many students can apply either formula in isolation yet stall when a problem hands them an instrument reading and expects them to notice it is someone else's variable. The workshop trains that recognition step, not new algebra.

Does every integrated problem in this workshop reduce to one of the three chains described here

The three chains cover the large majority of problems built from this syllabus's instruments chapter, because they pair each instrument type (ammeter, voltmeter and ammeter together, wattmeter) with the machine or circuit quantity it is designed to read. Occasionally a problem strings two chains together, such as using an OC-test voltmeter and ammeter to also check the magnetising branch — but the underlying method, trace the reading back to its formula, stays the same.

Integrated circuit and machine problem workshop

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads