I think I know the answer, but I'm unsure. I'm studying for my FE, and in the practice exam is the following question:

An ideal transformer connects a 120-V, 60-Hz voltage source to a load. One terminal of the voltage source is grounded. The load is ungrounded. The circuit and transformer are operating normally (not in a fault condition). A person whose body is contacting the ground is in proximity to the circuit. If the person touches the circuit, which point likey has the highest risk of resulting in shock?


simulate this circuit – Schematic created using CircuitLab

The correct answer is A. I can justify this in several ways, but what confused me was the explanation in the back of the practice exam: It says that point C and D have no differential voltage to ground.

Could someone please explain why this is the case?

My hypothesis (formed from half remembered lectures) is that because the transformer is inducing a current in the secondary coil, the current must flow through the coil and cannot flow through only part of it to jump to the ground.

  • 1
    \$\begingroup\$ This is a terribly dangerous question to put on the FE. It presents an ideal circuit, which is rarely reality. It has the possibility to give someone the impression that this is somehow a safe practice. There is always parasitic capacitance (winding to winding, and winding to ground) that may very well provide an adequate path for fatal current. \$\endgroup\$ Commented Jun 24, 2021 at 1:48

3 Answers 3


The transformer secondary current is fully contained in the load and in wires C and D. There is no loop formed to ground (ideal transformer, wires and load, no stray leakage.)

If a grounded person touches the secondary at a single point, that point will assume the GND voltage, but again since there's no loop between the (grounded) primary and secondary, no current will flow to GND or through the person (only by definition of an ‘ideal transformer.’ More about that below.) They could touch the secondary anywhere, so long as they touch it at single point.

Given one and only one point ground on the secondary, the rest of the secondary loop will be at different voltages than the contact point. If a second ground-referenced contact is made to one of those different voltages on the secondary, there will be current. But that's not the case being asked.

This quiz question notwithstanding, a real system with leakage and capacitance will actually have both a transient current when you touch the secondary as its capacitance is charged to the ground voltage, and a continuous leakage current coupling from primary to secondary.

There are regulations that set limits on this leakage to non-hazardous levels for consumer devices: 3mA for isolated low-voltage devices, like a USB power adapter.

Higher-voltage secondaries will have hazardous levels of leakage even if they're fully floating just because of their physical size, and so are usually grounded and/or insulated against accidental contact.

tl; dr: even a 'floating' secondary can kill you. Use caution.


It is not true that points C and D have no voltage with respect to ground. In fact, they may have a very high voltage with respect to ground. Since they have no conducting path to ground we simply can't say anything about the voltage between either of these points and ground.

However, there is no path from point C or D that can conduct current to ground. So, based only on the elements shown in your schematic (no capacitance) there can not be any current flow through your body if you contact ground and point C or D. It's the current that will kill you.


You are right to question the exam, there's a voltage from C to D they can't both be at ground potential. you can get that conclusion by applying kirchoff's loop law. \$ V_{CG} - V_{DG} = {VAB\over2} \$

If you were to build that circuit and wave a non-contact voltage dectctor pen over it, (or probe it with a sensitive multimeter) you would see a voltage from B and C to ground. (Also from A but its not interesting)

This voltage is just from capacitance to the mains supply if you were to proivide a low impedance path to ground only a very small current would flow.

  • \$\begingroup\$ Question specifies ‘ideal’ circuit - no leakage or parasitics. Which we know is a physical impossibility but that’s not the assumption. \$\endgroup\$ Commented Jun 24, 2021 at 5:36
  • \$\begingroup\$ the answer is in the first paragraph, the rest is just filler because the algorithm dislikes short answers. \$\endgroup\$ Commented Jun 24, 2021 at 22:00
  • \$\begingroup\$ That's the thing: by definition in the problem, there is no loop (complete isolation, no parasitics.) Therefore, no current. \$\endgroup\$ Commented Jun 24, 2021 at 22:00
  • \$\begingroup\$ the loop law does not require the node be connected in any way it still gives consistent results. complete isolation makes the voltage undefined, which is not anything like zero. \$\endgroup\$ Commented Jun 24, 2021 at 22:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.