One of the most common schemes for making a resistive shunt is the loop shunt. The idea is that by folding the resistive element so that the opposite currents are next to each other, their magnetic fields cancel, reducing the inductance. However, the forces on the shunt can be significant. The following is a drawing of such a shunt, with current I flowing from left to right. #### What's the resistance of this shunt, and how much power will be dissipated?

If we assume that d is 1 cm (about 1/2 inch), and L is 10 cm (4"), and we're using standard 1x1/8" bar stock, the resistance will be:

R = 1.71E-6 ohm cm * 21 cm/0.8 cm^2 = 0.045 mOhm.

At a current of 10 kA, the heat dissipated will be about 4.5 kW. A 10 microsecond pulse at that current will put about 0.045 Joules into the shunt, which is negligible.

#### What's the force F tending to pull the shunt apart?

Using Ampere's law, the force between two parallel conductors is: F = I*L/d, where F is in Newtons, I is in Amps, and Land d are in Meters (that is, 1 amp creates a force of 1 N between two wires 1 meter long). Plugging in some reasonable numbers, and assuming that the current carrying member is infinitely thin:

I = 10 kA, d = 0.01 m, L = 0.1 cm

F = 10000 * 0.1/0.01 = 100 kN, which is a substantial force (about 22,000 pounds).

If the shunt were made from copper bar that is 1"x1/8" (a typical size), let's figure out the shear load at point "P", assuming that the bar is perfectly rigid and doesn't flex to absorb some of the force. The cross sectional area will be 1/8 square inch, or about 0.8 sq cm, so the load is 100 kN/8E-5 = 1.24E9 N/sq meter (Pascal). The yield strength of hard copper is about 231 MPa (with ultimate strength of 351 MPa), although it varies somewhat with cold working, etc. (from http://www.matweb.com/ ) In any case, the force on the joint is more sufficient to break the copper, or, at the very least, to permanently deform the members.

loopshunt.htm / 26 Sep 2000 / Back to HV Home / Back to home page / Mail to Jim