2 Improved formatting edit approved Dec 10 '16 at 14:41 zx485 23011 silver badge66 bronze badges In this case you already represent both in the same terms (cosine) so you can see which one is leading by seeing which phase is greater. In this case v1(t) = 4cos(377t + 10)$$\v1(t) = 4*\cos(377t + 10)\$$ and v2 = +20cos (377t+180)$$\v2 = +20*cos (377t+180)\$$. You can see that v2 has greater phase (180) than v1 (10). Now you know that V2 is leading V1 by 170 (180-10). In this case you already represent both in the same terms (cosine) so you can see which one is leading by seeing which phase is greater. In this case v1(t) = 4cos(377t + 10) and v2 = +20cos (377t+180). You can see that v2 has greater phase (180) than v1 (10). Now you know that V2 is leading V1 by 170 (180-10). In this case you already represent both in the same terms (cosine) so you can see which one is leading by seeing which phase is greater. In this case $$\v1(t) = 4*\cos(377t + 10)\$$ and $$\v2 = +20*cos (377t+180)\$$. You can see that v2 has greater phase (180) than v1 (10). Now you know that V2 is leading V1 by 170 (180-10). 1 answered Dec 10 '16 at 13:52 shafiyyah 511 silver badge66 bronze badges In this case you already represent both in the same terms (cosine) so you can see which one is leading by seeing which phase is greater. In this case v1(t) = 4cos(377t + 10) and v2 = +20cos (377t+180). You can see that v2 has greater phase (180) than v1 (10). Now you know that V2 is leading V1 by 170 (180-10).