We calculate the 'maximum Zs' that will 'just' result in magnetic tripping of the MCB (in the 'worse case' for a B-Curve MCB requiring a current of 5 x In) with a supply voltage of 218.5V (230V x Cmin). i.e. , for a B-curve MCB ......
max Zs = (230 x Cmin) / (5 x In)
For any given (measured or calculated) Zs, the current during fault conditions (with a supply voltage of 218.5V) is therefore:
fault current = (230 x Cmin) / Zs
... for example, for a B32 MCB and a Zs of about 1.36Ω ...
fault current = (230 x Cmin) / 1.36 = 160.7 A
... i.e. just enough to trip a B32 (in 'worst case')
However, that assumes the the actual loop impedance of the fault circuit is equal to the (measured or calculated) Zs, and therefore assumes that the impedance of the actual fault is zero. If,with the same circuit, the fault actually had an impedance of, say, 0.5Ω, then we would have:
fault current = (230 x Cmin) / (1.36 + 0.5) = 117.5 A
... clearly not enough to trip a B32 in the worst case (magnetic trip threshold = 5 x In)
Hence, to get magnetic tripping in that circuit with a fault of just 0.5 impedance would require an MCB with a 'worst case' trip threshold of less than 117.5 A, maybe one with a spec requiring tripping at "2-3 x In".
Things obviously get 'worse' if the impedance of the fault is even greater (than 0.5Ω ). For example, if, with this circuit ,the fault had an impedance of 1Ω, then we would have:
fault current = (230 x Cmin) / (1.36 + 1.0) = 92,6 A
,... which would mean that we would need an MCB with a ('worst case') magnetic trip threshold below 92.6 A - maybe one with a spec requiring tripping at "1.0 - 2.5 x In" ...and so on, depending upon how high a fault impedance one wanted to guarantee would result in a magnetic trip.
Does that help?
Kind Regards, John