Fibrous tissue/scar does not conduct electricity (electricity has to go around it)
Problem - one of the routes is slower than normal (image - alpha is fast & beta is slow)
Alpha pathway has a longer refractory period than beta
When AP travels along both pathways, by the time the b pathway has sent its activity around the fibrous tissue, the a pathway has recovered and the b pathway ends up activating the a pathway (and going back up)
When AP has reached the top (C), the b pathway will have recovered, AP goes back round -> ends up producing a run of tachycardia