"i2x1" - 1 õppematerjal

ALALISVOOLU ELEKTRIAHELAD

U A 2 I 2 R2 I 3 R3 0 U A 2 I 2 R2 I 3 R3 Olgu UA1=4 V, UA2=5 V, R1=R2=R3=1 U A2 I 2 R2 I 3 R3 Kahe allikaga elektriahela arvutus paneme puuduva liikme asemele null-takistuse I1R1 + I2x0 + I3R3 = UA1 I1x0 + I2R2 + I3R3 = UA2 I1 + I 2 - I 3 = 0 Sisestame arvväärtused I II III IV I1x1 + I2x0 + I3x1 = UA1 I1x0 + I2x1 + I3x1 = UA2 I1x1 + I2x1 + I3x(-1) = 0  Koostame determinandid. 1 =1x1x(-1) + 0x1x1 + 1x0x1 ­ 1x1x1 ­ 1x1x1 ­ 0x0x(-1) = -1 +0 +0 ­1 ­1 +0 = -3 =4x1x(-1) + 0x1x0 + 1x5x1 ­ 1x1x0 ­ 4x1x1 ­ 0x5x(-1) = -4+0+5­0­4+5 = -3 Siis I1= 1/ = ­3/-3 = 1A  Koostame determinandid. 2 = 1x5x(-1) + 4x1x1 + 1x0x0 ­ 1x5x1 ­ 0x1x1 ­ 4x0x(-1) = -5+4+0 ­5­0+0 = -6 Siit I2=2/= -6/-3 = 2A =1x1x0 + 0x5x1 + 4x0x1 ­ 4x1x1 ­ 1x5x1 ­ 0x0x0 =0 +0 +0 ­4