<Dd> (3) − (I 3 − I 1) / 3, 0, (I 3 − I 1) / 3 (\ displaystyle - (I_ (3) - I_ (1)) / 3, 0, (I_ (3) - I_ (1)) / 3) </Dd> <P> The equivalence can be readily shown by using Kirchhoff's circuit laws that I 1 + I 2 + I 3 = 0 (\ displaystyle I_ (1) + I_ (2) + I_ (3) = 0). Now each problem is relatively simple, since it involves only one single ideal current source . To obtain exactly the same outcome voltages at the nodes for each problem, the equivalent resistances in the two circuits must be the same, this can be easily found by using the basic rules of series and parallel circuits: </P> <Dl> <Dd> R 3 + R 1 = (R c + R a) R b R a + R b + R c, (\ displaystyle R_ (3) + R_ (1) = (\ frac ((R_ (c) + R_ (a)) R_ (b)) (R_ (a) + R_ (b) + R_ (c))),) R 3 R 1 = R a R c . (\ displaystyle (\ frac (R_ (3)) (R_ (1))) = (\ frac (R_ (a)) (R_ (c))).) </Dd> </Dl> <Dd> R 3 + R 1 = (R c + R a) R b R a + R b + R c, (\ displaystyle R_ (3) + R_ (1) = (\ frac ((R_ (c) + R_ (a)) R_ (b)) (R_ (a) + R_ (b) + R_ (c))),) R 3 R 1 = R a R c . (\ displaystyle (\ frac (R_ (3)) (R_ (1))) = (\ frac (R_ (a)) (R_ (c))).) </Dd>

Where do we use star-delta and delta-star transformation