Instrument transformers (CTs and VTs) on the PE Power exam

Instrument transformer questions on the PE Power exam ask you to move a current, voltage, power or impedance between the primary circuit and a meter or relay, and to judge whether the transformer can drive its connected load. The NCEES handbook gives you little here: the ideal transformer ratio under Single-Phase Transformer Equivalent Circuits and the device-number table under Single-Line Diagrams. Burden and accuracy-class relations are not printed in it, so you carry them in. Installation rules sit in NEC Articles 250 and 450.

What you actually need to own

• CT ratio. A 600:5 CT has a ratio of 120. Secondary current is primary current divided by the ratio, and a relay tap times the ratio is the primary pickup. A CT sits on one phase conductor, so no √3 enters for wye-connected CTs. Delta-connected CTs put √3 times the CT secondary current into the leads.
• Summed secondaries. Convert each CT through its own ratio, then add only the secondaries that reach the instrument. A reversed CT subtracts.
• CT burden. Burden is the load connected to the secondary terminals: instruments plus leads. Convert volt-amperes to ohms at rated secondary current, Z = VA / I_s², which is VA/25 for a 5 A secondary. When the current returns on a second lead, lead resistance uses the loop, twice the one-way run.
• Relaying accuracy class. C400 means the CT delivers 400 V at its secondary terminals at 20 times rated secondary current while ratio error stays within 10%. For a 5 A secondary that is 400 V / 100 A = 4 Ω of external burden: relays, meters and leads together. On a lower tap of a multi-ratio CT, the available voltage falls with the fraction of turns in use.
• Metering accuracy class. In 0.3B0.2, the 0.3 is the accuracy class in percent and the 0.2 after the B is the standard burden in ohms. The class holds for connected burdens up to that value.
• VT ratio and connection. Secondary voltage is the voltage applied to the primary divided by the nameplate ratio. Trace both primary leads. A unit connected line to neutral sees V_LL/√3, and one connected line to line sees V_LL.
• VT burden. Devices across a VT secondary are in parallel, so their volt-amperes add. Lead current is I = S_total / V, and lead drop is that current times the loop resistance.
• Through both transformers. Meter multiplier = CT ratio × VT ratio. A secondary watt or watthour reading times the multiplier is the primary quantity. A distance relay sees Z_secondary = Z_primary × CT ratio / VT ratio.
• Differential zones. A differential zone runs from CT to CT, whatever the breaker positions.
• What the NEC adds. Sections 250.170 to 250.178 cover grounding of instrument transformer secondary circuits and cases. Article 450 excludes current transformers, and 450.3(C) covers voltage transformer protection.

Where people lose points

• Treating 400:5 as 400:1. The answer is off by a factor of five.
• Using one-way lead length. The secondary circuit is a loop. On wye-connected CTs a ground fault uses both leads, and a balanced three-phase fault only the one-way run.
• Mishandling the CT winding resistance. It is inside the CT, so it is not part of the burden compared with a rating. It does join the loop when the question asks for the voltage the CT must develop internally.
• Assuming a C-class CT is good to 20 times rated at any burden. The burden decides when it saturates.
• Reading a metering class backwards, taking the burden figure as the accuracy.
• Adding VT burdens as series impedances. They are in parallel.
• Applying √3 to a VT already marked line to line, or missing it on a line-to-neutral unit.
• Using one ratio in a meter multiplier. Both transformers scale the reading.
• Opening an energized CT secondary. A dangerous voltage appears at the open terminals. Short the secondary first.

A note on the codes

The exam supplies its codes, including the 2020 NEC and the 2017 NESC, and scores code answers against those editions. See the PE Power handbook and code books.

How to study it

Pick one feeder with an 800:5 CT and a 7,200:120 VT. Refer a load current, a relay pickup, a wattmeter reading and a line impedance through the ratios in both directions. Then add leads and instruments, total the CT burden in ohms and compare it with a C-class limit and a metering limit. Check your line-to-neutral voltages against three-phase circuits, and carry the relay pickups into overcurrent protection and coordination.

Part of the Measurement and Instrumentation area of the PE Power exam. → Practice it on CastorPrep